Question #bba18ecb - SAT Suite Question Bank

ID: bba18ecb

Modded SAT Question Bank by Abdullah Mallik

When the quadratic function $f$ is graphed in the xy-plane, where $y = f (x)$ , its vertex is $left parenthesis negative 3 comma 6 right parenthesis$ . One of the x-intercepts of this graph is $(- \frac{17}{4}, 0)$ . What is the other x-intercept of the graph?

$(- \frac{29}{4}, 0)$
$left parenthesis negative seven fourths comma 0 right parenthesis$
$left parenthesis five fourths comma 0 right parenthesis$
$left parenthesis StartFraction 17 Over 4 EndFraction comma 0 right parenthesis$

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Correct Answer: B

Rationale

Choice B is correct. Since the line of symmetry for the graph of a quadratic function contains the vertex of the graph, the x-coordinate of the vertex of the graph of $y = f (x)$ is the x-coordinate of the midpoint of its two x-intercepts. The midpoint of two points with x-coordinates $x 1$ and $x 2$ has x-coordinate $x_{m}$ , where $x_{m} = \frac{x_{1} + x_{2}}{2}$ . It′s given that the vertex is $left parenthesis negative 3 comma 6 right parenthesis$ and one of the x-intercepts is $(- \frac{17}{4}, 0)$ . Substituting $negative 3$ for $x_{m}$ and $- \frac{17}{4}$ for $x 1$ in the equation $x_{m} = \frac{x_{1} + x_{2}}{2}$ yields $- 3 = \frac{- \frac{17}{4} + x_{2}}{2}$ . Multiplying each side of this equation by $2$ yields $- 6 = - \frac{17}{4} + x_{2}$ . Adding $StartFraction 17 Over 4 EndFraction$ to each side of this equation yields $- \frac{7}{4} = x_{2}$ . Therefore, the other x-intercept is $left parenthesis negative seven fourths comma 0 right parenthesis$ .

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

Choice C 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