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

Sep 4, 2017

$\text{see explanation}$

#### Explanation:

$\text{for a parabola in standard form } f \left(x\right) = a {x}^{2} + b x + c$

$\text{the x-coordinate of the vertex is }$

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

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

$\text{with } a = - 1 , b = 2 , c = - 6$

$\Rightarrow {x}_{\textcolor{red}{\text{vertex}}} = - \frac{2}{- 2} = 1$

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

${y}_{\textcolor{red}{\text{vertex}}} = - {1}^{2} + 2 - 6 = - 5$

$\Rightarrow \textcolor{m a \ge n t a}{\text{vertex }} = \left(1 , - 5\right)$

$\textcolor{b l u e}{\text{Intercepts}}$

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

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

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

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

$\text{check the "color(blue)"discriminant}$

$\Delta = {b}^{2} - 4 a c = 4 - 24 = - 20$

$\Delta < 0 \Rightarrow \text{ equation has no real solutions}$

$\text{hence there are no x-intercepts}$
graph{-x^2+2x-6 [-16.02, 16.02, -8, 8.02]}