Question

Write an equation for the parabola that is congruent to y = 3x^2, has vertex (3,-5), and opens down.

A) y = -3(x - 3)^2 - 5
B) y = 3(x - 3)^2 - 5
C) y = -3(x + 3)^2 - 5
D) y = 3(x + 3)^2 - 5

Answer 1

y = 3(x - 3)² - 5

The correct answer is A.

Answer 2

The correct equation of the parabola is y = -3(x - 3)^2 - 5, as it is congruent to y = 3x^2, has a vertex at (3, -5), and opens downward.

Step-by-step explanation:

The student asks for the equation of a parabola that is congruent to y = 3x^2, has vertex at (3, -5), and opens downward. To write this equation, we need to use the vertex form of a parabola's equation, which is y = a(x - h)^2 + k, where (h, k) is the vertex of the parabola and 'a' determines the direction of the opening as well as the width of the parabola. Since the original parabola has an 'a' value of 3 and opens upward, to have a congruent parabola that opens downward, we'll need to use -3 as our 'a' value. Therefore, the correct equation with the vertex (3, -5) and opening downward is y = -3(x - 3)^2 - 5, which is option A.