Find the quadratic function f(x)=ax²+bx+c for each parabola described: 1. minimum value -8, x-intercepts 0 and 8 2. vertex (2, 12); x-intercepts -4 and 8 3. minimum value -6; zeros of ƒ are -1 and 5 4. y-intercept 2; x-intercepts -2 and 4

1. This function goes trough 3 points: ( 0,0),(8,0) and (4, -8) so we have to find f(0)=0, f(8)=0, f(4)= -8

f(0)= c=0✔️
f(8)=64 a +8b=0 divide by 4; 16a+2b=0 (1)
f(4)=16a+4b=-8 divide by 4; 4a+b= -2 (2)

From(2)
b=-4a-2
Substitute b in (1)
16 a+2( -4a-2)=0; 16a-8a-4=0; 8a=4
so a=1/2✔️
b=-2-2= -4✔️

The quadratic function is:
f(x)=(X^2)/2 -4x=0

2. This quadratic function goes trough points ( 2,12), (-4,0), (8,0)

The quadratic function is:
f(x)= (-x^2)/3. + 4x/3. +32/3

3. This quadratic function goes trough points ( 2, -6), (-1,0), (5,0)

The quadratic function is:
f(x)=(2x^2)/3. -8x/3. -10/3

4. This function goes trough points (0,2), (-2,0), (4,0)

The quadratic function is:
f(x)=(-x^2)/4. +x/2. +2