SAT Mathematics Practice Test 6

Since the shown parabola opens upward, the coefficient of ${\mathit{x}}^2$ in the equation $\mathit{y}=\mathit{a}{\mathit{x}}^2+\mathit{c}$ must be positive. Given that a is positive,$-\mathit{a}$ is negative, and therefore the graph of the equation $\mathit{y}=\mathit{a}{\left(\mathit{x}-\mathit{b}\right)}^2+\mathit{c}$ will be a parabola that opens downward. The vertex of this parabola is (b, c), because the maximum value of y, c, is reached when x = b. Therefore, the answer must be choice B.Choices A and C are incorrect. The coefficient of ${\mathit{x}}^2$ in the equation $\mathit{y}=-\mathit{a}{\left(\mathit{x}-\mathit{b}\right)}^2+\mathit{c}$ is negative.Therefore, the parabola with this equation opens downward, not upward. Choice D is incorrect because the vertex of this parabola is $\left(\mathit{b},\mathit{c}\right)$, not $\left(-\mathit{b},\mathit{c}\right),$ because the maximum value of $\mathit{y},\mathit{c},$ is reached when $\mathit{x}=\mathit{b}$.