Answer to Question #202374 in Algebra for platon

Question

a quadratic relation has zeros at -1 and 3, and a y-intercept of -18. Determine the equation of the relation in vertex form

Answer 1

We want quadratic in vertex form. In the vertex form quadratic equation is of type

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

given that quadratic has zeros at -1 and 3.

so quadratic equation must have factors $(x+1) and (x-3).$

so our quadratic will be of type$y=p(x+1)(x+3)$

now we want value of p.

given that quadratic relation has y intercept of -18.

so when $u=0, y=-18\\ ie 18=p{0+1}{0-3}\\ =-18=p(-3)\\ p=6\\$

so our quadratic relation is $y=6(x+1)(x-3)$

now

$y=6(x+1)(x-3)\\ y=(6x+6)(x-3)=(6x^2-18x+6x-18)\\ y=6x^2-12x-18\\ y=6(x^2-2x-3)\\ y=6(x^2-2x+1-1-3)\\ y=6(x-1)^1-4)\\ y=6(x-1)^2-24\\$

This is the required vertex form.