Question #a54753ca - SAT Suite Question Bank

ID: a54753ca

Modded SAT Question Bank by Abdullah Mallik

In the xy-plane, the graph of the equation $y = - x^{2} + 9 x - 100$ intersects the line $y = c$ at exactly one point. What is the value of $c$ ?

$- \frac{481}{4}$
$negative 100$
$- \frac{319}{4}$
$- \frac{9}{2}$

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Correct Answer: C

Rationale

Choice C is correct. In the xy-plane, the graph of the line $y = c$ is a horizontal line that crosses the y-axis at $y = c$ and the graph of the quadratic equation $y = - x^{2} + 9 x - 100$ is a parabola. A parabola can intersect a horizontal line at exactly one point only at its vertex. Therefore, the value of $c$ should be equal to the y-coordinate of the vertex of the graph of the given equation. For a quadratic equation in vertex form, $y = a {(x - h)}^{2} + k$ , the vertex of its graph in the xy-plane is $(h, k)$ . The given quadratic equation, $y = - x^{2} + 9 x - 100$ , can be rewritten as $y = - (x^{2} - 2 (\frac{9}{2}) x + {(\frac{9}{2})}^{2}) + {(\frac{9}{2})}^{2} - 100$ , or $y = - {(x - \frac{9}{2})}^{2} + (- \frac{319}{4})$ . Thus, the value of $c$ is equal to $- \frac{319}{4}$ .

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 D is incorrect and may result from conceptual or calculation errors.

Question Difficulty: Hard