Question #92f812bb - SAT Suite Question Bank

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Modded SAT Question Bank by Abdullah Mallik

In the xy-plane, a parabola has vertex $left parenthesis 9 comma negative 14 right parenthesis$ and intersects the x-axis at two points. If the equation of the parabola is written in the form $y = a x^{2} + b x + c$ , where $a$ , $b$ , and $c$ are constants, which of the following could be the value of $a + b + c$ ?

$negative 23$
$negative 19$
$negative 14$
$negative 12$

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Correct Answer: D

Rationale

Choice D is correct. The equation of a parabola in the xy-plane can be written in the form $y = a {(x - h)}^{2} + k$ , where $a$ is a constant and $(h, k)$ is the vertex of the parabola. If $a$ is positive, the parabola will open upward, and if $a$ is negative, the parabola will open downward. It’s given that the parabola has vertex $left parenthesis 9 comma negative 14 right parenthesis$ . Substituting $9$ for $h$ and $negative 14$ for $k$ in the equation $y = a {(x - h)}^{2} + k$ gives $y = a {(x - 9)}^{2} - 14$ , which can be rewritten as $y = a (x - 9) (x - 9) - 14$ , or $y = a (x^{2} - 18 x + 81) - 14$ . Distributing the factor of $a$ on the right-hand side of this equation yields $y = a x^{2} - 18 a x + 81 a - 14$ . Therefore, the equation of the parabola, $y = a x^{2} - 18 a x + 81 a - 14$ , can be written in the form $y = a x^{2} + b x + c$ , where $a = a$ , $b = - 18 a$ , and $c = 81 a - 14$ . Substituting $minus 18 a$ for $b$ and $81 a minus 14$ for $c$ in the expression $a + b + c$ yields $(a) + (- 18 a) + (81 a - 14)$ , or $64 a minus 14$ . Since the vertex of the parabola, $left parenthesis 9 comma negative 14 right parenthesis$ , is below the x-axis, and it’s given that the parabola intersects the x-axis at two points, the parabola must open upward. Therefore, the constant $a$ must have a positive value. Setting the expression $64 a minus 14$ equal to the value in choice D yields $64 a - 14 = - 12$ . Adding $14$ to both sides of this equation yields $64 a equals 2$ . Dividing both sides of this equation by $64$ yields $a = \frac{2}{64}$ , which is a positive value. Therefore, if the equation of the parabola is written in the form $y = a x^{2} + b x + c$ , where $a$ , $b$ , and $c$ are constants, the value of $a + b + c$ could be $negative 12$ .

Choice A is incorrect. If the equation of a parabola with a vertex at $left parenthesis 9 comma negative 14 right parenthesis$ is written in the form $y = a x^{2} + b x + c$ , where $a$ , $b$ , and $c$ are constants and $a + b + c = - 23$ , then the value of $a$ will be negative, which means the parabola will open downward, not upward, and will intersect the x-axis at zero points, not two points.

Choice B is incorrect. If the equation of a parabola with a vertex at $left parenthesis 9 comma negative 14 right parenthesis$ is written in the form $y = a x^{2} + b x + c$ , where $a$ , $b$ , and $c$ are constants and $a + b + c = - 19$ , then the value of $a$ will be negative, which means the parabola will open downward, not upward, and will intersect the x-axis at zero points, not two points.

Choice C is incorrect. If the equation of a parabola with a vertex at $left parenthesis 9 comma negative 14 right parenthesis$ is written in the form $y = a x^{2} + b x + c$ , where $a$ , $b$ , and $c$ are constants and $a + b + c = - 14$ , then the value of $a$ will be $0$ , which is inconsistent with the equation of a parabola.

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