Question

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

Answer 1

Vertex: `(-1/2, 5)`

`y` intercept: `(0, 4)`

`x` intercepts: `(1/2+-sqrt(5)/2, 0)`

Explanation:

Given:

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

We can complete the square to get this into vertex form:

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

`color(white)(f(x)) = -4(x^2-x-1)`

`color(white)(f(x)) = -4(x^2-x+1/4-5/4)`

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

`color(white)(f(x)) = -4(x-1/2)^2+5`

This is in vertex form:

`f(x) = a(x-h)^2+k`

where `a=-4` is the multiplier (affecting the steepness and up/down orientation of the parabola) and `(h,k) = (-1/2, 5)` is the vertex.

We can use the difference of squares identity:

`a^2-b^2 = (a-b)(a+b)`

with `a=2(x-1/2)` and `b=sqrt(5)` to get it into factored form so we can find the zeros:

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

`color(white)(f(x)) = -((2(x-1/2))^2-(sqrt(5))^2)`

`color(white)(f(x)) = -(2(x-1/2)-sqrt(5))(2(x-1/2)+sqrt(5))`

`color(white)(f(x)) = -(2x-1-sqrt(5))(2x-1+sqrt(5))`

Hence the zeros are:

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

So the `x` intercepts are:

`(1/2+sqrt(5)/2, 0)" "` and `" "(1/2 - sqrt(5)/2, 0)`

The `y` intercept is `f(0) = 0+0+4 = 4`, i.e. `(0, 4)`

graph{-4x^2+4x+4 [-4, 4, -3.12, 6.88]}

Answer 2

Vertex `(0.5, 5)`
Intercept `(0,4)`

Explanation:

Given -

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

y - intercept

At x = 0;

`y=-4(0)^2+4(0)+4=0`
`y=4`

Intercept `(0,4)`

Vertex

`x=(-b)/(2xxa)=(-4)/(2 xx(-4))=(-4)/(-8)=1/2=0.5`

At `x=0.5`

`y=-4(0.5)^2+4(0.5)+4=-1+2+4=5`

Vertex `(0.5, 5)`