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