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

Aug 7, 2017

Vertex: $\left(- \frac{1}{2} , 5\right)$

$y$ intercept: $\left(0 , 4\right)$

$x$ intercepts: $\left(\frac{1}{2} \pm \frac{\sqrt{5}}{2} , 0\right)$

#### Explanation:

Given:

$f \left(x\right) = - 4 {x}^{2} + 4 x + 4$

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

$f \left(x\right) = - 4 {x}^{2} + 4 x + 4$

$\textcolor{w h i t e}{f \left(x\right)} = - 4 \left({x}^{2} - x - 1\right)$

$\textcolor{w h i t e}{f \left(x\right)} = - 4 \left({x}^{2} - x + \frac{1}{4} - \frac{5}{4}\right)$

$\textcolor{w h i t e}{f \left(x\right)} = - 4 \left({\left(x - \frac{1}{2}\right)}^{2} - \frac{5}{4}\right)$

$\textcolor{w h i t e}{f \left(x\right)} = - 4 {\left(x - \frac{1}{2}\right)}^{2} + 5$

This is in vertex form:

$f \left(x\right) = a {\left(x - h\right)}^{2} + k$

where $a = - 4$ is the multiplier (affecting the steepness and up/down orientation of the parabola) and $\left(h , k\right) = \left(- \frac{1}{2} , 5\right)$ is the vertex.

We can use the difference of squares identity:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

with $a = 2 \left(x - \frac{1}{2}\right)$ and $b = \sqrt{5}$ to get it into factored form so we can find the zeros:

$f \left(x\right) = - 4 {\left(x - \frac{1}{2}\right)}^{2} + 5$

$\textcolor{w h i t e}{f \left(x\right)} = - \left({\left(2 \left(x - \frac{1}{2}\right)\right)}^{2} - {\left(\sqrt{5}\right)}^{2}\right)$

$\textcolor{w h i t e}{f \left(x\right)} = - \left(2 \left(x - \frac{1}{2}\right) - \sqrt{5}\right) \left(2 \left(x - \frac{1}{2}\right) + \sqrt{5}\right)$

$\textcolor{w h i t e}{f \left(x\right)} = - \left(2 x - 1 - \sqrt{5}\right) \left(2 x - 1 + \sqrt{5}\right)$

Hence the zeros are:

$x = \frac{1}{2} \left(1 \pm \sqrt{5}\right)$

So the $x$ intercepts are:

$\left(\frac{1}{2} + \frac{\sqrt{5}}{2} , 0\right) \text{ }$ and $\text{ } \left(\frac{1}{2} - \frac{\sqrt{5}}{2} , 0\right)$

The $y$ intercept is $f \left(0\right) = 0 + 0 + 4 = 4$, i.e. $\left(0 , 4\right)$

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

Aug 7, 2017

Vertex $\left(0.5 , 5\right)$
Intercept $\left(0 , 4\right)$

#### Explanation:

Given -

$f \left(x\right) = - 4 {x}^{2} + 4 x + 4$

y - intercept

At x = 0;

$y = - 4 {\left(0\right)}^{2} + 4 \left(0\right) + 4 = 0$
$y = 4$

Intercept $\left(0 , 4\right)$

Vertex

$x = \frac{- b}{2 \times a} = \frac{- 4}{2 \times \left(- 4\right)} = \frac{- 4}{- 8} = \frac{1}{2} = 0.5$

At $x = 0.5$

$y = - 4 {\left(0.5\right)}^{2} + 4 \left(0.5\right) + 4 = - 1 + 2 + 4 = 5$

Vertex $\left(0.5 , 5\right)$