Question

The vertex of a parabola is (-2, -20), and its y-intercept is (0, -12).

Answer 1

Answer:

$y=2x^2+8x-12$

Step-by-step explanation:

To write the quadratic equation, begin by writing it in vertex form  

$y = a(x-h)^2+k$

Where (h,k) is the vertex of the parabola.

Here the vertex is (-2,-20). Substitute and write:

$y=a(x--2)^2+-20\\y=a(x+2)^2-20$

To find a, substitute one point (x,y) from the parabola into the equation and solve for a. Plug in (0,-12) the y-intercept of the parabola.

$-12=a((0)+2)^2-20\\-12=a(2)^2-20\\-12=4a-20\\8=4a\\2=a$

The vertex form of the equation is $y=2(x+2)^2-20$.

You can convert this into standard form by using the distributive property.

$y=2(x+2)^2-20\\y=2(x^2+4x+4)-20\\y=2x^2+8x+8-20\\y=2x^2+8x-12$