SAT Quadratic Functions

To solve problems with quadratic equations on the SAT, you need to know:

• the standard form of a quadratic equation
• the geometric interpretation of the coefficients of a quadratic equation
• how to factor quadratics
• the Rules of Exponents

If $f(x)=ax^{2}+bx+c$ where $a$ and $c$ are constants such that $a<0$ and $c>0,$ which of the following could be the graph of $f$?

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Solution:

Tip: The parabola $y=ax^{2}+bx+c$ opens down if $a<0.$
Tip: The parabola $y=ax^{2}+bx+c$ has a $y-$intercept at $y=c.$
Because $a<0,$ the parabola opens downward, and we can eliminate choices (A) and (B) where this is not the case.

If we set $x=0$, then $f(0)=c$. Hence, $c$ is the y-intercept. We are told that $c>0$. In choice (D), $c=0,$ and in choice (E), $c<0.$ Therefore we can eliminate them.

Only the parabola in choice (C) satisfies both conditions, $a<0$ and $c>0.$ It is the correct answer.

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Incorrect Choices:

(A), (B), (D), and (E)
See the solution for why these choices are wrong.

The graphs of $y=x^{2}-2$ and $y=-x^{2}+h$ intersect at point $(-\sqrt{5}, p)$ where $p$ is a constant. What is the value of $h$?

(A) $\ \ 12$
(B) $\ \ 11$
(C) $\ \ 10$
(D) $\ \ 9$
(E) $\ \ 8$

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Solution 1:

Tip: If two functions, $f$ and $g$, intersect at point $(x,y),$ then $f(x) = y = g(x)$.
Since $y=x^{2}-2$ intersects $y=-x^{2}+h$ at $(-\sqrt{5}, p)$, then $x^{2}-2=p=-x^{2}-h.$ We solve for $h$:

$$\begin{array}{r c l l} x^{2}-2&=&-x^{2}+h &\quad \text{set equations equal to each other}\\ 2x^{2}-2&=&h &\quad \text{add}\ x^{2}\ \text{to both sides}\\ 2x^{2}&=&2+h &\quad \text{add}\ 2\ \text{to both sides}\\ 2(-\sqrt{5})^{2}&=&2+h &\quad \text{plug in}\ x=-\sqrt{5} \\ 2\cdot 5 &=& 2+h &\quad \text{simplify}\\ 10&=&2+h &\quad 2 \cdot 5 =10\\ 8&=&h &\quad \text{subtract}\ 2\ \text{from both sides}\\ \end{array}$$

Solution 2:

Tip: If two functions, $f$ and $g$, intersect at point $(x,y),$ then $f(x) = y = g(x)$.
We plug $x=-\sqrt{5}$ into $y=x^{2}-2:$

$$y=x^{2}-2=(-\sqrt{5})^{2}-2=5-2=3$$

Now we substitute $x$ with $-\sqrt{5}$ and $h$ with each of the answer choices in $y=-x^{2}+h.$ We look for the answer choice that yields $y=3.$

(A) If $h=12$, $y=-x^{2}+h=-(-\sqrt{5})^{2}+12=-5+12=7 \neq 3.$ Wrong choice.
(B) If $h=11$, $y=-x^{2}+h=-(-\sqrt{5})^{2}+11=-5+11=6 \neq 3.$ Wrong choice.
(C) If $h=10$, $y=-x^{2}+h=-(-\sqrt{5})^{2}+10=-5+10=5 \neq 3.$ Wrong choice.
(D) If $h=9$, $y=-x^{2}+h=-(-\sqrt{5})^{2}+9=-5+9=4 \neq 3.$ Wrong choice.
(E) If $h=8$, $y=-x^{2}+h=-(-\sqrt{5})^{2}+8=-5+8=3=3.$ This is the correct answer.

Solution 3:

Tip: Use a calculator.
We use a graphing calculator to graph $y=x^{2}-2$. For each answer choice, we graph $y=-x^{2}+h$ and find where it intersects $y=x^{2}-2.$ The intersection point that has an $x-$coordinate equal to $-\sqrt{5} \approx -2.236$ will have a $y-$coordinate equal to $h.$

Below, $y=x^{2}-2$ is graphed in green. We only show the parabola for the correct answer (E), for which $h=8$. It is graphed in orange.

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Incorrect Choices:

(A), (B), (C), and (D)
Solution 2 eliminates these choices by plugging and checking. Solution 3 eliminates them using a graphing calculator.

If you thought these examples difficult and you need to review the material, these links will help:

• Standard Form of a Quadratic Equation
• Geometric Interpretation of the Coefficients of a Quadratic Function
• Factoring Quadratics
• Rules of Exponents
  $a^m \times a^n = a^{ m + n }$, $a^n / a^m = a^ { n - m }\ldots$

• Know the Rules of Exponents.
• Follow order of operations.
• SAT General Tips