# How do you find the vertex and the intercepts for f(x)=-x^2+6x+6 ?

Sep 2, 2017

$\text{see explanation}$

#### Explanation:

$\text{for the quadratic equation in standard form}$

•color(white)(x)y=ax^2+bx+c color(white)(x);a!=0

${x}_{\textcolor{red}{\text{vertex}}} = - \frac{b}{2 a}$

$f \left(x\right) = - {x}^{2} + 6 x + 6 \text{ is in standard form}$

$\text{with } a = - 1 , b = 6 , c = 6$

$\Rightarrow {x}_{\textcolor{red}{\text{vertex}}} = - \frac{6}{- 2} = 3$

$\text{substitute this value into the equation for y}$

$y = f \left(3\right) = - {\left(3\right)}^{2} + 6 \left(3\right) + 6 = 15$

$\Rightarrow \textcolor{m a \ge n t a}{\text{vertex }} = \left(3 , 15\right)$

$\textcolor{b l u e}{\text{for intercepts}}$

• " let x = 0, in equation for y-intercept"

• " let y = 0, in equation for x-intercepts"

$x = 0 \to y = 6 \leftarrow \textcolor{red}{\text{ y-intercept}}$

$y = 0 \to - {x}^{2} + 6 x + 6 = 0$

$\text{solve for x using the "color(blue)"quadratic formula}$

$x = \frac{- 6 \pm \sqrt{36 + 24}}{- 2}$

$\textcolor{w h i t e}{x} = \frac{- 6 \pm \sqrt{60}}{- 2}$

$\textcolor{w h i t e}{x} = \frac{- 6 \pm 2 \sqrt{15}}{- 2} = 3 \pm \sqrt{15}$

$\Rightarrow x \approx - 0.87 , x \approx 6.87 \leftarrow \textcolor{red}{\text{ x-intercepts}}$