How do you find the vertex and the intercepts for Y=x^2-2x-6?

Sep 1, 2017

$\text{see explanation}$

Explanation:

$\text{given the equation of a parabola in standard form}$

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

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

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

$y = {x}^{2} - 2 x - 6 \text{ is in standard form}$

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

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

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

$\Rightarrow {y}_{\textcolor{red}{\text{vertex}}} = 1 - 2 - 6 = - 7$

$\Rightarrow \textcolor{m a \ge n t a}{\text{vertex }} = \left(1 , - 7\right)$

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

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

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

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

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

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

$x = \frac{2 \pm \sqrt{4 + 24}}{2}$

$\textcolor{w h i t e}{x} = \frac{2 \pm \sqrt{28}}{2}$

$\textcolor{w h i t e}{x} = \frac{2 \pm 2 \sqrt{7}}{2} = 1 \pm \sqrt{7}$

$\Rightarrow x \approx - 1.65 , x \approx 3.65 \leftarrow \textcolor{red}{\text{ x-intercepts}}$
graph{x^2-2x-6 [-20, 20, -10, 10]}