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

Aug 27, 2017

Vertex: $\left(2 , 8\right)$

X-intercepts: $\left(2 \left(1 + \sqrt{2}\right) , 0\right) ,$ $\left(2 \left(1 - \sqrt{2}\right) , 0\right)$

Approximate intercepts: $\left(- 0.828 , 0\right) ,$$\left(4.83 , 0\right)$

Y-Intercept: $\left(0 , 4\right)$

#### Explanation:

Given:

$f \left(x\right) = - {x}^{2} + 4 x + 4$ is a quadratic equation in standard form:

$f \left(x\right) = a {x}^{2} + b x + c$,

where:

$a = - 1$, $b = 4$, and $c = 4$.

Vertex: maximum or minimum point of a parabola

The $x$ value of the vertex is the same as the axis of symmetry that divides the parabola into two equal halves.

$x = \frac{- b}{2 a}$

$x = \frac{- 4}{2 \cdot - 1}$

$x = \frac{- 4}{- 2}$ $\leftarrow$ Two negatives make a positive.

$x = 2$

To determine the $y$ value of the vertex, substitute $y$ for $f \left(x\right)$. Substitute $2$ for $x$ and solve for $y$.

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

Simplify.

$y = - 4 + 8 + 4$

$y = 8$

The vertex is $\left(2 , 8\right)$.

Since $a < 0$, the vertex is the maximum point and the parabola opens downward.

X-Intercepts: value of $x$ when $y = 0$

Substitute $0$ for $f \left(x\right)$ and solve for $x$ using the quadratic formula.

$0 = - {x}^{2} + 4 x + 4$

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

Plug in the known values.

$x = \frac{- 4 \pm \sqrt{{\left(4\right)}^{2} - 4 \cdot - 1 \cdot 4}}{2 \cdot - 1}$

Simplify.

$x = \frac{- 4 \pm \sqrt{16 + 16}}{- 2}$

Simplify $16 + 16$ to $32$.

$x = \frac{- 4 \pm \sqrt{32}}{- 2}$

Prime factorize $32$.

$x = \frac{- 4 \pm \sqrt{\left(2 \times 2\right) \times \left(2 \times 2\right) \times 2}}{- 2}$

Simplify.

$x = \frac{- 4 \pm 2 \times 2 \sqrt{2}}{- 2}$

$x = \frac{- 4 \pm 4 \sqrt{2}}{- 2}$

Factor out the common $2$.

x=(color(red)cancel(color(black)(-4))^2+-color(red)cancel(color(black)(4))^2sqrt2)/(color(red)cancel(color(black)(-2))^1

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

Solutions for $x$.

$x = 2 + 2 \sqrt{2} ,$$2 - 2 \sqrt{2}$

Simplify.

$x = 2 \left(1 + \sqrt{2}\right) ,$$2 \left(1 - \sqrt{2}\right)$

x-intercepts: (2(1+sqrt2)),0), (2(1-sqrt2)),0)

Approximate intercepts: $\left(- 0.828 , 0\right) ,$$\left(4.83 , 0\right)$

Y-Intercept: value of $y$ when $x = 0$.

Substitute $0$ for $x$ and solve for $y$.

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

$y = 4$

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

Plot the points and sketch a parabola through the dots. Do not connect the dots.

graph{y=-x^2+4x+4 [-16.02, 16.02, -8.01, 8.01]}