Question

A parabola that has a vertex of (3, -2) a focus of (3, -2 1/16) , and opens downward. Which of the following best represents the equation of the parabola in standard form?

Answer 1

Answer:

$y=-8x^2+48x-74$

Step-by-step explanation:

A parabola that has a vertex of (3, -2) a focus of (3, -2 1/16), then the line of symmetry is x = 3.

The distance between the vertex and focus is equal to p/2, so

$\dfrac{p}{2}=\sqrt{(3-3)^2+\left(-2+2 \dfrac{1}{16}\right)^2}=\dfrac{1}{16},$

so parabola's equation in vertex form is

$(x-x_0)^2=-2p(y-y_0)\\ \\(x-3)^2=-2\cdot \dfrac{1}{16}\cdot (y+2)\\ \\(x-3)^2=-\dfrac{1}{8}(y+2)\\ \\8(x-3)^2=-(y+2)$

In standard form this equation is

$8(x^2-6x+9)=-y-2\\ \\y=-8x^2+48x-72-2\\ \\y=-8x^2+48x-74$