Question

Create the equation of a quadratic function, in standard form, that has one zero of -3 and a turning point at (-1,-16)

Answer 1

Answer:

$y=4x^2+8x-12$

Step-by-step explanation:

we know that

The quadratic equation in vertex form is equal to

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

where

a is the leading coefficient

(h,k) is the vertex of the quadratic equation

Remember that

In a quadratic equation the turning point is the same that the vertex

so

(h,k)=(-1,-16)

substitute

$y=a(x+1)^2-16$

we have one zero at (-3,0)

substitute and solve for a

$0=a(-3+1)^2-16$

$0=4a-16\\4a=16\\a=4$

substitute

$y=4(x+1)^2-16$

Convert to standard form

$y=4(x^2+2x+1)-16\\y=4x^2+8x+4-16\\y=4x^2+8x-12$