Help with quadratic function problem?

2(x^2+2x-3)=y

x=-3

(x+3)=0

x=1

(x-1)=0

(x+3)(x-1)=0

(x^2+2x-3)=y Times by 2.

2(x^2+2x-3)=y

Answer is 2x^2+4x-6=y

If -3 is an x-intercept, then the function has (x+3) as a factor (plugging x=-3 will give f(-3)=0).

Another factor is (x-1) for our other intercept x=1.

Multiplying these two factors gives a coefficient of 1 for x² so you need to double it to get a=2.

Answer: f(x) = 2(x+3)(x-1) = 2x² + 4x - 6.

The answer is 2(x--3)(x-1)=2x^2+4x-6.

you would take x=-3 and x=1

so y=(x+3)(x-1)

then y=x^2 -x +3x -3

y=(x^2)+2x-3

but we need a to equal 2 so we times it all by 2 and get

y=2(x^2)+4x-6

y=2x^2 + 4X - 6

you have to use the quadratic equation:

x= the opposite of b plus or minus radical (square root) b squared minus 4 ac over 2a. (hey!)

a is the first number so you plug that in to the 4ac and 2a.

the a is a 2??? *boggles*

as in the y = Ax^2 + Bx + C .... the A = 2?

then (-B +/- SQRT(B^2 - 4AC))/2A = -3 and 1

B = 4

SQRT(16 - 8C) = 2

16 - 8C = 4

C = 3/2

y = 2x^2 + 4X + 3/2