Question #4dd4efcf - SAT Suite Question Bank

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Modded SAT Question Bank by Abdullah Mallik

$f (x) = a x^{2} + 4 x + c$

In the given quadratic function, $a$ and $c$ are constants. The graph of $y = f (x)$ in the xy-plane is a parabola that opens upward and has a vertex at the point $(h, k)$ , where $h$ and $k$ are constants. If $k less than 0$ and $f (- 9) = f (3)$ , which of the following must be true?

$c less than 0$
$a greater than or equals 1$

I only
II only
I and II
Neither I nor II

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Correct Answer: D

Rationale

Choice D is correct. It's given that the graph of $y = f (x)$ in the xy-plane is a parabola with vertex $(h, k)$ . If $f (- 9) = f (3)$ , then for the graph of $y = f (x)$ , the point with an x-coordinate of $negative 9$ and the point with an x-coordinate of $3$ have the same y-coordinate. In the xy-plane, a parabola is a symmetric graph such that when two points have the same y-coordinate, these points are equidistant from the vertex, and the x-coordinate of the vertex is halfway between the x-coordinates of these two points. Therefore, for the graph of $y = f (x)$ , the points with x-coordinates $negative 9$ and $3$ are equidistant from the vertex, $(h, k)$ , and $h$ is halfway between $negative 9$ and $3$ . The value that is halfway between $negative 9$ and $3$ is $\frac{- 9 + 3}{2}$ , or $negative 3$ . Therefore, $h = - 3$ . The equation defining $f$ can also be written in vertex form, $f (x) = a {(x - h)}^{2} + k$ . Substituting $negative 3$ for $h$ in this equation yields $f (x) = a {(x - (- 3))}^{2} + k$ , or $f (x) = a {(x + 3)}^{2} + k$ . This equation is equivalent to $f (x) = a (x^{2} + 6 x + 9) + k$ , or $f (x) = a x^{2} + 6 a x + 9 a + k$ . Since $f (x) = a x^{2} + 4 x + c$ , it follows that $6 a equals 4$ and $9 a + k = c$ . Dividing both sides of the equation $6 a equals 4$ by $6$ yields $a = \frac{4}{6}$ , or $a = \frac{2}{3}$ . Since $two thirds less than 1$ , it's not true that $a greater than or equals 1$ . Therefore, statement II isn't true. Substituting $\frac{2}{3}$ for $a$ in the equation $9 a + k = c$ yields $9 (\frac{2}{3}) + k = c$ , or $6 + k = c$ . Subtracting $6$ from both sides of this equation yields $k = c - 6$ . If $k less than 0$ , then $c minus 6 less than 0$ , or $c less than 6$ . Since $c$ could be any value less than $6$ , it's not necessarily true that $c less than 0$ . Therefore, statement I isn't necessarily true. Thus, neither I nor II must be true.

Choice A is incorrect and may result from conceptual or calculation errors.

Choice B is incorrect and may result from conceptual or calculation errors.

Choice C is incorrect and may result from conceptual or calculation errors.

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