Precalculus Mathematics 1 Study Guide

When it comes to college courses, Precalculus Mathematics 1 is a course that you will have to understand how to do the following things to get ready for the actual final exam of this respective mathematics course:

• Finding the domain of functions
• Finding the intercepts of functions
• Finding the (horizontal and vertical) asymptotes of functions
• Finding the average rate of change
• Finding inverse functions
• Finding the difference quotient of functions
• Expanding logarithmic expressions
• Solving logarithmic and exponential equations
• Graphing functions
• Solving polynomic and rational inequalities
• Convert logarithms into exponential form
• Constructing polynomials
• Evaluating functions
• Solving functions
• Finding the zeros of polynomic and rational functions
• Function composition
• Computing functions using function composition
• Finding values of logarithms
• Finding the equation of an ellipse
• Exponential growth
• Dividing and factorizing polynomic expressions
• Finding the equation of a parabola

P.S. I am writing this study guide based on Barry University's MAT 109 final exam study guide that I was given. However, if you would like to see the full document of Barry University's MAT 109 final exam study guide, then you can check my version of it on either Academia, CourseHero, or even on my LinkedIn. Click here if you want the Precalculus Mathematics 2 Study Guide.

When it comes to finding the domain of various functions, you have to know what type of function you are dealing with in order to how to set and find its domain. \[\begin{array}{ll} \textbf{Function}&\textbf{Domain}\\ \text{Constant}& \text{Domain}=\Re \\ \text{Polynomial}& \text{Domain }=\Re \\ \text{Rational}& \text{Set denominator } \neq 0 \\ \text{Radical}&\text{Set inside } \ge 0 \\ \text{Logarithmic} & \text{Set inside } > 0 \\ \end{array}\]

Find the domain of the following function: \[f(x)=\frac{2x+3}{x^2-5x+6}\]

Reveal the answer

When it comes to finding the domain of a rational function, you have to set the denominator unequal to \(0\) (since you are finding the undefined, or unknown, points of the function), which means that you are going to set \(x^2-5x+6\) to \(0\), which means doing the following: \[\begin{matrix} &x^2-5x+6\neq 0&\\ &x^2-2x-3x+6\neq 0&\\ &\left(x^2-2x\right)+\left(-3x+6\right)\neq 0&\\ &x\left(x-2\right)-3\left(x-2\right)\neq 0&\\ &\left(x-2\right)\left(x-3\right)\neq 0&\\ x-2=0&&x-3\neq 0\\ x-2+2\neq 0+2&&x-3+3\neq 0+3\\ x\neq 2&&x\neq 3\\ &\boxed{\mathbf{{\color{red} \{x\mid x\neq 2,3\} }}}& \end{matrix}\]

Find the domain of the following function: \[f(x)=\sqrt{2x-6}\]

Reveal the answer

When it comes to finding the domain of a square root function, you have to set the interior of the radical greater than or equal to \(0\), which means doing the following: \[\begin{matrix} 2x-6\ge 0\\ 2x-6+6\ge 0+6\\ 2x\ge 6\\ \frac{2x}{2}\ge \frac{6}{2}\\ x\ge 3\\ \boxed{\mathbf{{\color{red} \{x\mid x\ge 3\} }}} \end{matrix}\]

Find the domain of the following function: \[f(x)=\frac{1}{\sqrt{2x-6}}\]

Reveal the answer

When it comes to finding the domain of a rational function, you have to set the denominator greater than \(0\), which means that you are going to set \(\sqrt{2x-6}\) to \(0\), which is also ironic that it's literally the same concept as the previous problem, but the only difference is that you have an extra step to work on (along with using \(>\) instead of \(\ge\) thanks to the denominator being a radical), which means doing the following: \[\begin{matrix} \sqrt{2x-6}> 0\\ (\sqrt{2x-6})^2> (0)^2\\ 2x-6> 0\\ 2x-6+6> 0+6\\ 2x> 6\\ \frac{2x}{2}> \frac{6}{2}\\ x> 3\\ \boxed{\mathbf{{\color{red} \{x\mid x> 3\} }}} \end{matrix}\]

Find the domain of the following function: \[f(x)=\log_3\left(x^2-6x+8\right)\]

Reveal the answer

When it comes to finding the domain of a logarithmic function, you have to set the interior of the logarithmic function \(>\) \(0\), which means doing the following: \[\begin{matrix} &x^2-6x+8>0&\\ &x^2-2x-4x+8>0&\\ &\left(x^2-2x\right)+\left(-4x+8\right)>0&\\ &x\left(x-2\right)-4\left(x-2\right)>0&\\ &\left(x-2\right)\left(x-4\right)>0&\\ x-2<0&&x-4>0\\ x-2+2<0+2&&x-4+4>0+4\\ x<2&&x>4\\ &\boxed{\mathbf{{\color{red} \{x\mid x<2\:or\:x>4\} }}}& \end{matrix}\]

Find the domain of the following function: \[f(x)=\frac{1}{x(x^2+4)}\]

Reveal the answer

When it comes to finding the domain of a rational function, you have to set the denominator unequal to \(0\), which means that you are going to set \(x(x^2+4)\) to \(0\), which means doing the following: \[\begin{matrix} &x(x^2+4)\neq 0&\\ x\neq 0&&x^2+4\neq 0\\ &&x^2+4-4\neq 0-4\\ &&x^2\neq -4\\ &&\sqrt{x^2}\neq \sqrt{-4}\\ &&x\neq \pm\sqrt{-4}\\ &&\textup{Undefined}\\ &\boxed{\mathbf{{\color{red} \{x\mid x\neq 0\} }}}& \end{matrix}\]

The domain was already covered in Section A, but to find the domain of the rational function, then you just have to set the denominator to \(0\).

When it comes to finding the intercepts, the best thing to do is to let \(y=0\) in order to find the \(x\)-intercepts while letting \(x=0\) in order to find the \(y\)-intercepts.

When it comes to finding a rational function's vertical asymptote, you set the denominator equal to \(0\) then solve for \(x\). When it comes to finding a rational function's horizontal asymptote, you divide the numerator's leading coefficient by the denominator's leading coefficient.

Find the intercepts of the following rational function: \[f(x)=\frac{2x-36}{5x-60}\]

Reveal the answer

\[\begin{matrix} &f(x)=\frac{2x-36}{5x-60}&\\ \textup{\textbf{x-intercept}}&&\textup{\textbf{y-intercept}}\\ \frac{2x-36}{5x-60}=0&&f(0)=\frac{2(0)-36}{5(0)-60}\\ \frac{2x-36}{5x-60}(5x-60)=0(5x+35)&&f(0)=\frac{0-36}{0-60}\\ 2x-36=0&&f(0)=\frac{36}{60}\\ 2x=0 + 36&&y=\frac{36\div12}{60\div12}\\ 2x=36&&y = \frac35 \\ \frac{2x}{2} = \frac{36}{2}&& \\ x=18&&\\\\ (x,0)&&(0,y)\\ \boxed{\mathbf{{\color{red} (18,0) }}}&&\boxed{\mathbf{{\color{red} \left ( 0,\frac{3}{5} \right ) }}} \end{matrix}\]

Find the asymptotes of the following rational function: \[f(x)=\frac{3x+24}{5x+35}\]

Reveal the answer

\[\begin{matrix} \textup{\textbf{Vertical Asymptote}}\\ 5x+35= 0\\ 5x+35-35= 0-35\\ 5x= -35\\ \frac{5x}{5}=-\frac{35}{5}\\ \boxed{\mathbf{{\color{red} x= -7 }}}\\\\ \textup{\textbf{Horizontal Asymptote}}\\ y=\frac{3x+24}{5x+35}\rightarrow \frac{3x}{5x}=\frac{3}{5}\rightarrow \boxed{\mathbf{{\color{red} y=\frac{3}{5} }}} \end{matrix}\]

When it comes to the average rate of change, it is best to know what the formula for the average rate of change is, which is \[A(x)=\frac{f(b)-f(a)}{b-a}\] with \(A(x)\) being the average rate of change along with \(b\) being the higher term and \(a\) being the lower term. The funny thing about the finding the average rate of change of a linear function is commonly going to be the linear function's slope.

Find the average rate of change the following function from \(2\) to \(5\): \[f(x)=5x+4\]

Reveal the answer

\[\begin{matrix} A(x)=\frac{f(b)-f(a)}{b-a}=\frac{f\left(5\right)-f\left(2\right)}{5-2}=\frac{\left(5\left(5\right)+4\right)-\left(5\left(2\right)+4\right)}{3}=\frac{\left(25+4\right)-\left(10+4\right)}{3}=\frac{\left(29\right)-\left(14\right)}{3}=\frac{15}{3}=\boxed{\mathbf{{\color{red} 5 }}} \end{matrix}\]

Find the average rate of change the following function from \(-3\) to \(1\): \[f(x)=2x+1\]

Reveal the answer

\[\begin{matrix} \begin{matrix} A(x)=\frac{f(b)-f(a)}{b-a}=\frac{f\left(1\right)-f\left(-3\right)}{1-(-3)}=\frac{(2(1)+1)-(2(-3)+1)}{1+3}=\frac{(2+1)-(-6+1)}{4}=\frac{(3)-(-5)}{4}=\frac{(3)+5}{4}=\frac{8}{4}=\boxed{\mathbf{{\color{red} 2 }}} \end{matrix} \end{matrix}\]

Find the average rate of change the following function from \(1\) to \(x\): \[f(x)=3x-4\]

Reveal the answer

\[\begin{matrix} A(x)=\frac{f(b)-f(a)}{b-a}=\frac{f\left(x\right)-f\left(1\right)}{x-1}=\frac{(3(x)-4)-(3(1)-4)}{x-1}=\frac{(3x-4)-(3-4)}{x-1}=\frac{(3x-4)-(-1)}{x-1}=\frac{(3x-4)+1}{x-1}=\frac{3x+3}{x-1}=\frac{3(x-1)}{x-1}=\boxed{\mathbf{{\color{red} 3 }}} \end{matrix}\]

When it comes to the finding the inverse of a linear function, it is best to note that you change \(x\) with \(y\) then solve for \(y\).

\[\begin{matrix} \textup{\textbf{If}}&\textup{\textbf{Then}}\\ f(x)=mx+b&f^{-1}\left(x\right)=\frac{x-b}{m}\\\\ f(x)=mx-b&f^{-1}\left(x\right)=\frac{x+b}{m} \end{matrix}\]

Find the inverse, \(f^{-1}\left(x\right)\) of the following function: \[f(x)=3x+5\]

Reveal the answer

\[\begin{matrix} f\left(x\right)=3x+5\\ y=3x+5\\ x=3y+5\\ x-5=3y+5-5\\ x-5=3y\\ \frac{x-5}{3}=\frac{3y}{3}\\ \frac{x-5}{3}=y\\ \boxed{\mathbf{{\color{red} y=\frac{x-5}{3} }}} \end{matrix}\]

Find the inverse, \(f^{-1}\left(x\right)\) of the following function: \[f(x)=\frac{x-4}{7}\]

Reveal the answer

\[\begin{matrix} f\left(x\right)=\frac{x-4}{7}\\ y=\frac{x-4}{7}\\ x=\frac{y-4}{7}\\ x(7)=\frac{y-4}{7}(7)\\ 7x=y-4\\ 7x+4=y-4+4\\ 7x+4=y\\ \boxed{\mathbf{{\color{red} y=7x+4 }}} \end{matrix}\]

Find the Difference Quotient \(\left[\left(D.Q\right)=\frac{f\left(x+h\right)-f\left(x\right)}{h}\right]\) for each \(f(x)\).

When it comes to the difference quotient, you're basically making sure that you plug in \(x+h\) instead of \(x\) for the first part of the equation hence why the equation is literally \[\frac{f(x+h)-f(x)}{h}.\]

Find the difference quotient of the following function: \[f(x)=5x+4\]

Reveal the answer

\[\begin{matrix} \frac{f\left(x+h\right)-f\left(x\right)}{h}&=\frac{(5\left(x+h\right)+4)-(5x+4)}{h}\\\\ &=\frac{(5x+5h+4)-(5x+4)}{h}\\\\ &=\frac{5x+5h+4-5x-4}{h}\\\\ &=\frac{5x-5x+5h+4-4}{h}\\\\ &=\frac{(5-5)x+5h+(4-4)}{h}\\\\ &=\frac{(0)x+5h+(0)}{h}\\\\ &=\frac{5h}{h}\\\\ &=\boxed{\mathbf{{\color{red} 5 }}} \end{matrix}\]

Find the difference quotient of the following function: \[f(x)=3x^2+1\]

Reveal the answer

\[\begin{matrix} \frac{f\left(x+h\right)-f\left(x\right)}{h}&=\frac{(3\left(x+h\right)^2+1)-(3x^2+1)}{h}\\\\ &=\frac{(3(x^2+2xh+h^2)+1)-(3x^2+1)}{h}\\\\ &=\frac{3x^2+6xh+3h^2+1-3x^2-1}{h}\\\\ &=\frac{3x^2-3x^2+6xh+3h^2+1-1}{h}\\\\ &=\frac{(3-3)x^2+6xh+3h^2+(1-1)}{h}\\\\ &=\frac{(0)x^2+6xh+3h^2+(0)}{h}\\\\ &=\frac{6xh+3h^2}{h}\\\\ &=\frac{h(6x+3h)}{h}\\\\ &=\boxed{\mathbf{{\color{red} 6x+3h }}} \end{matrix}\]

When it comes to expanding logarithmic expressions, you have to make sure you know the basic logarithmic rules, which some are the following: \[\begin{matrix} \log(ab)=\log\:a+\log\:b\\\\ \log(\frac{a}{b})=\log\:a-\log\:b\\\\ \log(a^n)=n\:\log\:a \end{matrix}\]

Write the following logarithmic as a sum/difference of logarithms: \[\log(a^2b^5c^6)\]

Reveal the answer

\[\begin{matrix} \log\left(a^2b^5c^6\right)&=\log\left(a^2\right)+\log\left(b^5\right)+\log\left(c^6\right)\\ &=\boxed{\mathbf{{\color{red} 2\log\left(a\right)+5\log\left(b\right)+6\log\left(c\right) }}} \end{matrix}\]

Write the following logarithmic as a sum/difference of logarithms: \[\log(5x^3y)\]

Reveal the answer

\[\begin{matrix} \log\left(5x^3y\right)&=\log\left(5\right)+\log\left(x^3\right)+\log\left(y\right)\\ &=\boxed{\mathbf{{\color{red} \log\left(5\right)+3\log\left(x\right)+\log\left(y\right) }}} \end{matrix}\]

Write the following logarithmic as a sum/difference of logarithms: \[\log\left [ \frac{a^2b^3}{c^4d^5} \right ]\]

Reveal the answer

\[\begin{matrix} \log\left[\frac{a^2b^3}{c^4d^5}\right]&=\log(a^2b^3)-\log(c^4d^5)\\ &=\log(a^2b^3)-\log(c^4d^5)\\ &=\log(a^2)+\log(b^3)-\log(c^4)+\log(d^5)\\ &=\boxed{\mathbf{{\color{red} 2\log(a)+3\log(b)-4\log(c)+5\log(d) }}} \end{matrix}\]

When it comes to solving each logarithmic equation, there are 2 different methods towards solving this equation. One method can be using the logarithmic-exponential conversion, which is \[\textup{If} \log_a\left(b\right)=c, \textup{then}\: a^c=b\] for the constant term then rewriting the logarithmic equation, which points out that \[\textup{If} \log_a\left(b\right)=\log_a\left(c\right), \textup{then}\: b=c.\] Another method can be using the general logarithmic rule (the first one explained for this section) then solving the logarithmic equation as a regular equation.

Solve the following logarithmic equation: \[\log_3\left(4x-7\right)=2\]

Reveal the answer

\[\begin{matrix} &\log _3\left(4x-7\right)=2&\\ \textup{\textbf{Method 1}}&&\textup{\textbf{Method 2}}\\ \log _3\left(4x-7\right)=2&&\log _3\left(4x-7\right)=2\\ \textup{Turn 2 into a log}&&3^2=4x-7\\ \log _3\left(x\right)=2&&9=4x-7\\ 3^2=x&&9+7=4x-7+7\\ 9=x&&16=4x\\ \therefore \log _3\left(9\right)=2&&\frac{16}{4x}=\frac{4}{4x}\\ \log _3\left(4x-7\right)=\log _3\left(9\right)&&\boxed{\mathbf{{\color{red} x=4 }}}\\ 4x-7=9&&\\ 4x-7+7=9+7&&\\ 4x=16&&\\ \frac{4}{4x}=\frac{16}{4x}&&\\ \boxed{\mathbf{{\color{red} x=4 }}}&& \end{matrix}\]

Solve the following logarithmic equation: \[\log_2\left(x-3\right)+\log_2\left(x-4\right)=1\]

Reveal the answer

\[\begin{matrix} \textup{\textbf{Method 1}}&\textup{\textbf{Method 2}}\\ \log_2\left(x-3\right)+\log_2\left(x-4\right)=1&\log_2\left(x-3\right)+\log_2\left(x-4\right)=1\\ \textup{Turn 1 into a log}&\log_2\left(x-3\right)\left(x-4\right)=1\\ \log_2\left(x\right)=1&2^1=\left(x-3\right)\left(x-4\right)\\ 2^1=x&2=\left(x-3\right)\left(x-4\right)\\ x=2&2=x^2-7x+12\\ \therefore \log _2\left(2\right)=1&2-2=x^2-7x+12-2\\ \log_2\left(x-3\right)+\log_2\left(x-4\right)=\log_2\left(2\right)&0=x^2-7x+10\\ \log_2\left(x-3\right)\left(x-4\right)=\log_2\left(2\right)&0=x^2-2x-5x+10\\ \left(x-3\right)\left(x-4\right)=2&0=\left(x^2-2x\right)+\left(-5x+10\right)\\ x^2-7x+12=2&0=x\left(x-2\right)-5\left(x-2\right)\\ x^2-7x+12-2=2-2&0=\left(x-2\right)\left(x-5\right)\\ x^2-7x+10=0&x=2,5\\ x^2-2x-5x+10=0&\\ \left(x^2-2x\right)+\left(-5x+10\right)=0&\\ x\left(x-2\right)-5\left(x-2\right)=0&\\ \left(x-2\right)\left(x-5\right)=0&\\ x=2,5& \end{matrix}\]

However, when it comes to this one, you can now verify both solutions while making sure they are true, which can be done by doing the following: \[\begin{matrix} \textup{\textbf{x=2}}&\textup{\textbf{x=5}}\\ \log_2\left(x-3\right)+\log_2\left(x-4\right)=1&\log_2\left(x-3\right)+\log_2\left(x-4\right)=1\\ \log_2\left((2)-3\right)+\log_2\left((2)-4\right)=1&\log_2\left((5)-3\right)+\log_2\left((5)-4\right)=1\\ \log_2\left(-1\right)+\log_2\left(-2\right)=1&\log_2\left(2\right)+\log_2\left(1\right)=1\\ \textup{Undefined}=1&\log_2\left(2\right)\left(1\right)=1\\ \textup{Undefined}\neq 1&2^1=\left(2\right)\left(1\right)\\ &2=2\\ &\textup{\textbf{True}}\\ \end{matrix}\]

The reason that \(\log_2\left(-1\right)+\log_2\left(-2\right)\) is labeled as "undefined" because you can't have a negative logarithm, which means that the final answer is only \(\boxed{\mathbf{{\color{red} x=5 }}}\).

When it comes to solving exponential equations, make sure that the bases are the same then solve for the exponents as a regular equation.

Solve the following exponential equation: \[9^{3x+2}=27^{x-5}\]

Reveal the answer

\[\begin{matrix} &9^{3x+2}=27^{x-5}&\\ 9=3^2&&27=3^3\\ &\left(3^2\right)^{3x+2}=\left(3^3\right)^{x-5}&\\ &3^{2\left(3x+2\right)}=3^{3\left(x-5\right)}&\\ &\textup{If}\: a^b=a^c, \textup{then}\: b=c.&\\ &2\left(3x+2\right)=3\left(x-5\right)&\\ &6x+4=3x-15&\\ &6x+4-4=3x-15-4&\\ &6x=3x-19&\\ &6x-3x=3x-19-3x&\\ &3x=-19&\\ &\frac{3x}{3}=\frac{-19}{3}&\\ &\boxed{\mathbf{{\color{red} x=-\frac{19}{3} }}}& \end{matrix}\]

When it comes to graphing each function, it is quite self-explanatory, but one good way to graph them is create a table then graph the points, which you can then connect the points together.

Graph the following function: \[f(x)=(x-5)^2+3\]

Reveal the answer

Here is a little table of plotting the points: \[\begin{matrix} & f(x)=(x-5)^2+3 & \\ x & &y \\ 2 & ((2)-5)^2+3 &12 \\ 4 & ((4)-5)^2+3 &4 \\ 5 & ((5)-5)^2+3 &3 \\ 6 & ((6)-5)^2+3 &4 \\ 8 & ((8)-5)^2+3 &12 \end{matrix}\]

Here is a graph with some of the plotted points:

When it comes to polynomic inequalities, you can just factor (if needed) then solve. When it comes to rational inequalities, you can just find the restrictions based on the denominator then solve for the given variable.

Solve the following polynomic inequality: \[x^2-8x+12\ge 0\]

Reveal the answer

\[\begin{matrix} &x^2-8x+12\ge 0&\\ &x^2-2x-6x+12\ge 0&\\ &\left(x^2-2x\right)+\left(-6x+12\right)\ge 0&\\ &x\left(x-2\right)-6\left(x-2\right)\ge 0&\\ &\left(x-2\right)\left(x-6\right)\ge 0&\\ x-2\le 0&\textup{or}&x-6\ge 0\\ x-2+2\le 0+2&\textup{or}&x-6+6\ge 0+6\\ \boxed{\mathbf{{\color{red} x\le 2 }}}&\boxed{\mathbf{{\color{red} \textup{or} }}}&\boxed{\mathbf{{\color{red} x\ge 6 }}} \end{matrix}\]

Solve the following rational inequality: \[\frac{x-2}{x-5}\ge 0\]

Reveal the answer

\[\begin{matrix} &\frac{x-2}{x-5}\ge 0&\\ &(x-2)^{1}(x-5)^{-1}\ge 0&\\ (x-2)^{1}\le 0&\textup{or}&(x-5)^{-1}\ge 0\\ x-2\le 0&\textup{or}&((x-5)^{-1})^{-1}\ge (0)^{-1}\\ x-2+2\le 0+2&\textup{or}&x-5\ge 0\\ x\le 2&\textup{or}&x-5+5\ge 0+5\\ &\textup{or}&x\ge 5\\\\ \boxed{\mathbf{{\color{red} x\le 2 }}}&\boxed{\mathbf{{\color{red} \textup{or} }}}&\boxed{\mathbf{{\color{red} x\ge 5 }}} \end{matrix}\]

When it comes to converting logarithmic equations into exponential form, then you are basically making sure that you are using the following logarithmic rule: \[\begin{matrix} \textup{\textbf{If}}&\textup{\textbf{Then}}\\ \log_a(b)=c&a^c=b\\ \ln_e(b)=c&e^c=b\\ \end{matrix}\]

Convert the following logarithmic equation into exponential form: \[\log_232=5\]

Reveal the answer

\[\begin{matrix} \log_232=5\\ \boxed{\mathbf{{\color{red} 2^5=32 }}} \end{matrix}\]

Convert the following logarithmic equation into exponential form: \[\ln(4)=1.4\]

Reveal the answer

\[\begin{matrix} \ln(4)=1.4\\ \ln_e(4)=1.4\\ \boxed{\mathbf{{\color{red} e^{1.4}=4 }}} \end{matrix}\]

When it comes to constructing a polynomial while given various specification, then you would have to understand the following when it comes to each part of the specifications:

• LC: the leading coefficient
• Degree: How many degrees are in the function
• Zero: Solve for 0 and that will be a factor
• Multiplicity: The exponent for the given factor

Construct a polynomial with the following specifications. Keep your answer in product form.

L.C. = \(2\), degree \(3\), zero \(-5\) with multiplicity \(2\) and zero \(1\) with multiplicity \(1\)

Reveal the answer

The \(\textup{\textbf{L.C.}}\) is the leading coefficient of the polynomial. Since L.C. = \(2\), then [so far] \[f(x)=2.\]

The \(\textup{\textbf{degree}}\) indicates that there will be, at most, a certain number of factors in this polynomial. Since degree is \(3\), then that means that will be, at least, \(3\) different factors in this polynomials.

The \(\textup{\textbf{zero}}\) indicates that, when solved for \(0\), you will receive the factors. Since, you are given "zero \(-5\)", and "zero \(1\)", then that means the following: \[\begin{matrix} \textup{zero}=-5&\textup{zero}=1\\\\ x=-5&x=1\\ x+5=-5+5&x-1=1-1\\ x+5=0&x-1=0\\\\ (x+5)&(x-1)\\ \end{matrix}\]

Therefore, this now makes the polynomial \[f(x)=2(x+5)(x-1).\]

The \(\textup{\textbf{multiplicity}}\) is the exponent of a given factor, which help indicate how many times a certain factor is being multiplied in the polynomial. Keep in mind that the degree plays a part of the multiplicity, the first term says "multiplicity \(2\)" then "multiplicity \(1\)". \(2\) multiplicities \(+\) \(1\) multiplicity equals \(3\) multiplicities, which means that, when simplified, will have a degree of \(3\). So, with that being said, the multiplicities can be detailed by the following: \[\begin{matrix} \textup{zero -5 with multiplicity 2}&\textup{zero 1 with multiplicity 1}\\\\ n=2&n=1\\ (x+5)^n&(x-1)^n\\ =(x+5)^2&=(x-1)^1\\ =(x+5)^2&=(x-1)\\ \end{matrix}\]

With that being said, then the constructed polynomial, in "product form", is \[\boxed{\mathbf{{\color{red} f(x)=2(x+5)^2(x-1) }}}\].

When it comes to evaluating functions when given various values, you are basically just substituting the values of \(x\) into the function then simplifying.

Evaluate the following function for given values of \(x\):

\(f(x)=5x^2+8\), for \(\sqrt{3}\), \(\sqrt{5}\), and \(\sqrt{7}\)

Reveal the answer

\[\begin{matrix} &&f\left(x\right)=5x^2+8&&\\ x=\sqrt{3}&&x=\sqrt{5}&&x=\sqrt{7}\\\\ \begin{matrix} f\left(x\right)&=&5x^2+8\\ f\left(\sqrt{3}\right)&=&5(\sqrt{3})^2+8\\ &=&5(3)+8\\ &=&15+8\\ &=&\boxed{\mathbf{{\color{red} 23 }}}\\ \end{matrix}&&\begin{matrix} f\left(x\right)&=&5x^2+8\\ f\left(\sqrt{5}\right)&=&5(\sqrt{5})^2+8\\ &=&5(5)+8\\ &=&25+8\\ &=&\boxed{\mathbf{{\color{red} 33 }}}\\ \end{matrix}&&\begin{matrix} f\left(x\right)&=&5x^2+8\\ f\left(\sqrt{7}\right)&=&5(\sqrt{7})^2+8\\ &=&5(7)+8\\ &=&35+8\\ &=&\boxed{\mathbf{{\color{red} 43 }}}\\ \end{matrix} \end{matrix}\]

If given a function then given a constant function, then set the two functions to be equivalent to each other and then solve for \(x\).

If \(f(x)=12(x-5)^2\), for what value of \(x\) would the following be:

\[f(x)=27\]

Reveal the answer

\[\begin{matrix} f(x)=12(x-5)^2&&f(x)=27\\ &y_1=y_2&\\ &12(x-5)^2=27&\\ &\frac{12\left(x-5\right)^2}{12}=\frac{27}{12}&\\ &\left(x-5\right)^2=\frac{9}{4}&\\ &\sqrt{\left(x-5\right)^2}=\sqrt{\frac{9}{4}}&\\\\ &x-5=\pm\frac{3}{2}&\\\\ &x-5+5=\pm\frac{3}{2}+5&\\\\ &x=5\pm\frac{3}{2}&\\ x=5-\frac{3}{2}&&x=5+\frac{3}{2}\\ x=\frac{10}{2}-\frac{3}{2}&&x=\frac{10}{2}+\frac{3}{2}\\ x=\frac{7}{2}&&x=\frac{13}{2}\\ &\boxed{\mathbf{{\color{red} x=\frac{7}{2},\:\frac{13}{2} }}}& \end{matrix}\]

This is somewhat the same concept as Section J. When it comes to finding the zeros of a function, set the function equal to \(0\) and solve for \(x\).

Find the zeros of the following polynomic function: \[f(x)=x^2-10x+16\]

Reveal the answer

\[\begin{matrix} &x^2-10x+16=0&\\ &x^2-2x-8x+16=0&\\ &\left(x^2-2x\right)+\left(-8x+16\right)=0&\\ &x\left(x-2\right)-8\left(x-2\right)=0&\\ &\left(x-2\right)\left(x-8\right)=0&\\ x-2=0&&x-8=0\\ x-2+2=0+2&&x-8+8=0+8\\ x=2&&x=8\\ &\boxed{\mathbf{{\color{red} x=2,8 }}}& \end{matrix}\]

Find the zeros of the following rational function: \[f(x)=\frac{5x^2-5}{x^2-5x+6}\]

Reveal the answer

\[\begin{matrix} &&\frac{5x^2-5}{x^2-5x+6}=0&&\\ &&\frac{5x^2-5}{x^2-5x+6}(x^2-5x+6)=0(x^2-5x+6)&&\\ &&5x^2-5=0&&\\ &&5(x^2-1)=0&&\\ 5=0&&&&x^2-1=0\\ 5\neq 0&&&&x^2-1^2=0\\ &&&&(x+1)(x-1)=0\\ &&&&\boxed{\mathbf{{\color{red} x=-1,1 }}} \end{matrix}\]

When it comes to function composition, they are basically "functions inside of functions". If you are given some functions then asked to find a certain composition, this means that you are substituting the "\(x\)" in \(f(x)\) as with \(g(x)\), which can be expressed by the following: \[\begin{matrix} (f\circ g)(x)=f(g(x))&(f\circ f)(x)=f(f(x))\\ (g\circ f)(x)=g(f(x))&(g\circ g)(x)=g(g(x))\\ \end{matrix}\]

For example (for the problem in this section), if \(f\left(x\right)=2x+3\) and \(g(x)=4x+1\), then \(f(g(x))=f(4x+1)\).

Find \((f\circ g)(x)\), \((g\circ f)(x)\), \((f\circ f)(x)\), and \((g\circ g)(x)\) if the following functions are given: \[\begin{matrix} f(x)=2x+3&g(x)=4x+1 \end{matrix}\]

Reveal the answer

\[\begin{matrix} f\left(x\right)=2x+3&&g\left(x\right)=4x+1\\\\ (f\circ g)(x)&&(f\circ f)(x)\\ \begin{matrix} f(x)&=&2x+3\\ f(g(x))&=&2(g(x))+3\\ &=&2(4x+1)+3\\ &=&8x+2+3\\ &=&8x+(2+3)\\ &=&8x+(5)\\ &=&\boxed{\mathbf{{\color{red} 8x+5 }}} \end{matrix}&&\begin{matrix} f(x)&=&2x+3\\ f(f(x))&=&2(f(x))+3\\ &=&2(2x+3)+3\\ &=&4x+6+3\\ &=&4x+(6+3)\\ &=&4x+(9)\\ &=&\boxed{\mathbf{{\color{red} 4x+9 }}} \end{matrix}\\ \end{matrix}\]

\[\begin{matrix} f\left(x\right)=2x+3&&g\left(x\right)=4x+1\\\\ (g\circ f)(x)&&(g\circ g)(x)\\ \begin{matrix} g(x)&=&4x+1\\ g(f(x))&=&4(f(x))+1\\ &=&4(2x+3)+1\\ &=&8x+12+1\\ &=&8x+(12+1)\\ &=&8x+(13)\\ &=&\boxed{\mathbf{{\color{red} 8x+13 }}} \end{matrix}&&\begin{matrix} g(x)&=&4x+1\\ g(g(x))&=&4(g(x))+1\\ &=&4(4x+1)+1\\ &=&16x+4+1\\ &=&16x+(4+1)\\ &=&16x+5\\ &=&\boxed{\mathbf{{\color{red} 16x+5 }}} \end{matrix} \end{matrix}\]

This is the same concept as Section P, but the only difference is that you are replacing \(x\) with a constant term.

Find \((f\circ g)(3)\), \((g\circ f)(2)\), \((f\circ f)(-3)\), and \((g\circ g)(5)\) if the following functions are given: \[\begin{matrix} f(x)=2x+3&g(x)=4x+1 \end{matrix}\]

Reveal the answer

\[\begin{matrix} f\left(x\right)=2x+3&&g\left(x\right)=4x+1\\\\ (f\circ g)(3)&&(f\circ f)(-3)\\ \begin{matrix} f(x)&=&2x+3\\ f(g(x))&=&2(g(x))+3\\ &=&2(4x+1)+3\\ &=&8x+2+3\\ &=&8x+(2+3)\\ &=&8x+(5)\\ &=&8x+5\\ f(g(3))&=&8(3)+5\\ &=&24+5\\ &=&\boxed{\mathbf{{\color{red} 29 }}} \end{matrix}&&\begin{matrix} f(x)&=&2x+3\\ f(f(x))&=&2(f(x))+3\\ &=&2(2x+3)+3\\ &=&4x+6+3\\ &=&4x+(6+3)\\ &=&4x+(9)\\ &=&4x+9\\ f(f(-3))&=&4(-3)+9\\ &=&-12+9\\ &=&\boxed{\mathbf{{\color{red} -3 }}}\\ \end{matrix} \end{matrix}\]

\[\begin{matrix} f\left(x\right)=2x+3&&g\left(x\right)=4x+1\\\\ (g\circ f)(2)&&(g\circ g)(5)\\ \begin{matrix} g(x)&=&4x+1\\ g(f(x))&=&4(f(x))+1\\ &=&4(2x+3)+1\\ &=&8x+12+1\\ &=&8x+(12+1)\\ &=&8x+(13)\\ &=&8x+13\\ g(f(2))&=&8(2)+13\\ &=&16+13\\ &=&\boxed{\mathbf{{\color{red} 29 }}}\\ \end{matrix}&&\begin{matrix} g(x)&=&4x+1\\ g(g(x))&=&4(g(x))+1\\ &=&4(4x+1)+1\\ &=&16+4+1\\ &=&16x+(5+1)\\ &=&16x+(6)\\ &=&16x+6\\ g(g(5))&=&16(5)+6\\ &=&80+6\\ &=&\boxed{\mathbf{{\color{red} 86 }}} \end{matrix} \end{matrix}\]

When it comes to computing piecewise functions, you are being asked to find the value of a function at a certain point, but now you're doing the same thing, but with a piecewise function instead. When dealing with a piecewise function, this is technically an introduction to intervals. When computing, you commonly are asked to find the sum or difference. You can be asked to find the product or quotient, but it's somewhat rare when you're being asked that.

Find \(f(-5)+f(3)\) and \(f(-5)-f(2)\) if the following functions are given: \[f(x)=\left\{\begin{matrix} 3x+5 & \textup{when}& x<0\\ 8 & \textup{when}& x=0\\ 5x+4 & \textup{when}& x>0 \end{matrix}\right.\]

Reveal the answer

Here is a good way to note which you need to know which intervals you should be dealing with: \[\left\{\begin{matrix} 3x+5 & \textup{when}& x<0\\ 8 & \textup{when}& x=0\\ 5x+4 & \textup{when}& x>0 \end{matrix}\right. \rightarrow \left\{\begin{matrix} -5<0 & & \\ & & \\ 2>0 & \textup{and} & 3>0 \end{matrix}\right. \]

Now that we know which intervals to use, you can now do substitution by doing the following: \[\begin{matrix} f(-5)+f(3)&&f(-5)-f(2)\\ (3x+5)+(5x+4)&&(3x+5)-(5x+4)\\ =(3(-5)+5)+(5(3)+4)&&=(3(-5)+5)-(5(2)+4)\\ =(-15+5)+(15+4)&&=(-15+5)-(10+4)\\ =(-10)+(19)&&=(-10)-(14)\\ =\boxed{\mathbf{{\color{red} 9 }}}&&=\boxed{\mathbf{{\color{red} -24 }}} \end{matrix}\]

This is basically the same concept as Section F where you first need to expand the logarithmic expression as a sum and/or difference of logarithms while using the same rules from Section F, which again are the following: \[\begin{matrix} \log(ab)=\log\:a+\log\:b\\\\ \log(\frac{a}{b})=\log\:a-\log\:b\\\\ \log(a^n)=n\:\log\:a \end{matrix}.\] Afterwards, this is basically allowing you to plug and play, or substituting, a variable for a constant then simplifying.

If \(log_b\:A=2\), \(log_b\:B=-1\), and \(log_b\:C=5\), then find the values of the following: \[\log_b\sqrt[3]{AB}\]

Reveal the answer

\[\begin{matrix} \log_b\sqrt[3]{AB}&=\log_b(AB)^{\frac{1}{3}}\\ &=\log_b(A)^{\frac{1}{3}}+\log_b(B)^{\frac{1}{3}}\\ &=\frac{1}{3}\log_b(A)+\frac{1}{3}\log_b(B)\\ &=\frac{1}{3}(2)+\frac{1}{3}(-1)\\ &=\frac{2}{3}-\frac{1}{3}\\ &=\boxed{\mathbf{{\color{red} \frac{1}{3} }}} \end{matrix}\]

When it comes to finding the equation of the ellipse with center at the origin, you're given the focus and vertex so just find the equation. There are two different formulas (the horizontal or even the vertical) to possibly use when it comes to the formula: \[\begin{matrix} \textup{\textbf{Horizontal}}&\textup{\textbf{Vertical}}\\ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1&\frac{y^2}{a^2}+\frac{x^2}{b^2}=1 \end{matrix}\]

The following is a fair way to explain the major axis when dealing with the determining whether is you are going to using either the horizontal or vertical formula: \[\begin{matrix} \textup{\textbf{Horizontal Major Axis}}&\textup{\textbf{Vertical Major Axis}}\\ \textup{Focus}=(\pm b,0)&\textup{Focus}=(0,\pm b)\\ \textup{Vertex}=(\pm a,0)&\textup{Vertex}=(0,\pm a) \end{matrix}\]

Find the equation of the ellipse with center at origin along with the following:

a focus at \((3,0)\) and a vertex at \((-4,0)\)

Reveal the answer

Based on these points, you have to use the horizontal version of this formula, which is \[\frac{x^2}{a^2}+\frac{y^2}{b^2}=1.\]

So, with that being said, you now have to note the following: \[\left\{\begin{matrix} a=-4 \\ b=3\\ (h,k)=(0,0) \end{matrix}\right.\]

Now, with all of that info listed, you can do the following in order to find out the equation of the ellipse: \[\begin{matrix} \frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1\\\\ \frac{(x-(0))^2}{(-4)^2}+\frac{(y-(0))^2}{(3)^2}=1\\\\ \frac{(x-0)^2}{16}+\frac{(y-0)^2}{9}=1\\\\ \boxed{\mathbf{{\color{red} \frac{x^2}{16}+\frac{y^2}{9}=1 }}} \end{matrix}\]

When dealing with (continuous) exponential growth and decay, you are basically noting that it's the \(r\) of the function is based on whether you are going to be dealing with exponential growth or decay. Anyways, the regular formula for exponential (growth or decay) would be the following: \[A=Pe^{rt}\: \textup{where} \left\{\begin{matrix} A=\textup{ending value}\\ P=\textup{initial value}\\ e=\textup{exponential}\approx 2.718\\ r=\textup{continuous rate}\\ t=\textup{time} \end{matrix}\right.\]

When it comes to the rate of this function, the following can determine whether you're dealing with exponential growth or decay: \[A=Pe^{rt}\: \textup{where} \left\{\begin{matrix} r<0&\rightarrow &\textup{exponential decay}\\ r>0&\rightarrow &\textup{exponential growth} \end{matrix}\right.\]

Here are other variations of this same formula: \[\begin{matrix} A=Pe^{rt}&A=A_0e^{kt}&N(t)=N_0e^{kt} \end{matrix}\]

A colony of bacteria grows according to the law of exponential growth \[(N\left(t\right)=N_0e^{0.17t}.)\]

If the initial number of bacteria was \(1000\), then find the time when it will be \(2500\).

Reveal the answer

So, "the initial number of bacteria was \(1000\)" means that \(N_0\) and \(2500\) will be \(N(t)\), which leaves only \(t\) as the only variable left in order to find since we have the time, which means doing the following in order to find the time: \[\begin{matrix} N\left(t\right)=N_0e^{0.17t}\\ (2500)=(1000)e^{0.17t}\\ 2500=1000e^{0.17t}\\ \frac{2500}{1000}=\frac{1000e^{0.17t}}{1000}\\ \frac{5}{2}=e^{0.17t}\\ \ln \left(\frac{5}{2}\right)=\ln \left(e^{0.17t}\right)\\ \ln \left(\frac{5}{2}\right)=0.17t\ln \left(e\right)\\ \ln \left(\frac{5}{2}\right)=0.17t\\ \frac{\ln \left(\frac{5}{2}\right)}{0.17}=\frac{0.17t}{0.17}\\ t=\frac{\ln \left(\frac{5}{2}\right)}{0.17}\\ \boxed{\mathbf{{\color{red} t\approx 5.39\: \textup{seconds} }}} \end{matrix}\]

When it comes to diving a polynomial by a polynomial, long division is highly required to use in order to solve them. However, make sure that you "add" and skipped terms before dividing just to be safe. When there is a remainder that's not \(0\), then the quotient become a "partial" quotient. If the degree of a remainder is less than the degree of the divisor, then that means you are done solving.

Find the quotient and the remained when the following is given: \[\begin{matrix} 5x^3-x-5&\textup{is divided by}&(x-1) \end{matrix}\]

Reveal the answer

Since the degree of the remainder, which is \(0\) since \(-1=0x-1=0x^0-1\) and that the remainder is a constant term, is less than the degree of the divisor, which is \(1\) since \(x-1=x^1-1\), then this means that you are done using long division. Therefore, this means that the following are the "partial" quotient and remainder: \[\begin{matrix} \textup{\textbf{Partial Quotient}}&\textup{\textbf{Remainder}}&&\textup{\textbf{Full Quotient}}\\ \boxed{\mathbf{{\color{red} 5x^2+5x+4 }}}&\boxed{\mathbf{{\color{red} -1 }}}&&5x^2+5x+4-\frac{1}{x-1} \end{matrix}\]

Keep in mind that the remainder is the constant term, which is \(-1\), and not the fraction, which is \(-\frac{1}{x-1}\) since the fraction will be part of the "full quotient".

There are 2 different formulas for the equation of a parabola, based on the directix you're given, which can be expressed by the following standard forms: \[\begin{matrix} \textup{\textbf{Horizontal}}&\textup{\textbf{Vertical}}\\ (x-h)^2=4p(y-k)&(y-k)^2=4p(x-h) \end{matrix}\]

When asked for the equation with the center at the origin, then the 2 formulas can be expressed as the following: \[\begin{matrix} \textup{\textbf{Horizontal}}&\textup{\textbf{Vertical}}\\ x^2=4py&y^2=4px \end{matrix}\]

This can be explained a little more by doing the following: \[\begin{matrix} \textup{\textbf{Axis of Symmetry}}&\textup{\textbf{Equation}}&\textup{\textbf{Directix}}&\textup{\textbf{Focus}}\\ \textup{Y-axis}&x^2=4py&y=-p&(0,p)\\ \textup{X-axis}&y^2=4px&x=-p&(p-0) \end{matrix}\]

Find the equation of the parabola with center at origin and the directix is the following line: \[y=-5\]

Reveal the answer

If \(y=-5\), then this means that the focus is \((5,0)\), which also means that \(p=5\). Since \(y=-5\), which means that you will be using the horizontal version of the equation. Now, with all of that being said, this means that you can do the following (both the standard and origin forms of the horizontal equation): \[\begin{matrix} \textup{\textbf{Standard Form}}&\textup{vs.}&\textup{\textbf{Origin Form}}\\ (x-h)^2=4p(y-k)&&x^2=4py\\ (x-(0))^2=4(5)(y-(0))&&x^2=4(5)(y)\\ (x)^2=20(y)&&x^2=20(y)\\ \boxed{\mathbf{{\color{red} x^2=20y }}}&&\boxed{\mathbf{{\color{red} x^2=20y }}}\\ \end{matrix}\]

Find the equation of the parabola with center at origin and the directix is the following line: \[x=-3\]

Reveal the answer

If \(x=-3\), then this means that the focus is \((3,0)\), which also means that \(p=3\). Since \(x=-3\), which means that you will be using the vertical version of the equation. Now, with all of that being said, this means that you can do the following (both the standard and origin forms of the vertical equation): \[\begin{matrix} \textup{\textbf{Standard Form}}&\textup{vs.}&\textup{\textbf{Origin Form}}\\ (y-k)^2=4p(x-h)&&y^2=4px\\ (y-(0))^2=4(3)(x-(0))&&y^2=4(3)(x)\\ (y)^2=12(x)&&y^2=12(x)\\ \boxed{\mathbf{{\color{red} y^2=12x }}}&&\boxed{\mathbf{{\color{red} y^2=12x }}}\\ \end{matrix}\]