# How do you find the vertex and intercepts for y = x^2 +2x - 5?

Nov 24, 2015

See explanation.

#### Explanation:

Finding the vertex:
Use the formula for the point of the vertex $\left(- \frac{b}{2 a} , f \left(- \frac{b}{2 a}\right)\right)$.

Remember that the quadratic is in the general form $a {x}^{2} + b x + c$, so $a = 1$ and $b = 2$.

$- \frac{b}{2 a} = - 1$

$f \left(- \frac{b}{2 a}\right) = f \left(- 1\right) = {\left(- 1\right)}^{2} + 2 \left(- 1\right) - 5 = - 6$

The vertex is at the point $\left(- 1 , - 6\right)$.

Finding the y-intercept:

Plug in $0$ for $x$.

$y = {\left(0\right)}^{2} + 2 \left(0\right) - 5 = - 5$

The $y$-intercept is $\left(0 , - 5\right)$.

Finding the x-intercept:

Plug in $0$ for $y$.

$0 = {x}^{2} + 2 x - 5$

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

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

$x = \frac{- 2 \pm 2 \sqrt{6}}{2} = - 1 \pm \sqrt{6}$

The $x$-intercepts are $\left(- 1 + \sqrt{6} , 0\right)$ and $\left(- 1 - \sqrt{6} , 0\right)$.