Question

Which equation when graphed has X-intercepts at (8,0) and (-2, 0) and a y-intercept at (0-48)?

Answer 1

Answer:

$y=3x^{2}-18x-48$

Step-by-step explanation:

we know that

The equation of a vertical parabola in factored form is equal to

$y=a(x-x1)(x-x2)$

where

a is a coefficient

x1 an x2 are the roots or x-intercepts

In this problem we have

$x1=8,x2=-2$

substitute

$y=a(x-8)(x+2)$

with the y-intercept (0,-48) find the value of a

substitute

$-48=a(0-8)(0+2)$

$-48=-16a$

$a=3$

The equation is equal to

$y=3(x-8)(x+2)\\ \\y=3(x^{2}+2x-8x-16)\\ \\ y=3(x^{2} -6x-16)\\ \\y=3x^{2}-18x-48$