Question

How do you find the vertex and intercepts for `f(x)= -4x^2 + 4x + 4`?

Answer 1

Vertex `(-1/2, 5)`
Y - intercept `(0, 4)`
X-intercept `( (sqrt5+1)/2,0); ((-sqrt5+1)/2,0)`

Explanation:

Given -

`f(x)=-4x^2+4x+4`

`y=-4x^2+4x+4`

Vertex

`x=(-b)/(2a)=(-(4))/(2xx-4)=(-4)/-8=1/2`

At `x=-1/2; y=-4(1/2)^2+4(1/2)+4=-1+2+4=5`

`(1/2, 5)`

At `x=0;y=-4(0)^2+4(0)+4=4`

Y - intercept `(0, 4)`

X-intercept

At `y=0; -4x^2+4x+4=0`

`-4x^2+4x+4=0`

Dividing both sides by `-4`

`x^2-x-1=0`

`x^2-x=1`

`x^2-x+1/4=1+1/4=5/4`

`(x-1/2)^2=5/4)`

`x-1/2=+-sqrt(5/4)=+-sqrt5/2`

`x=sqrt5/2+1/2=(sqrt5+1)/2`

`x=-sqrt5/2+1/2=(-sqrt5+1)/2`