How do you find the vertex and intercepts for y=x^2+4x-5?

Feb 25, 2016

Vertex: $\left(- 2 , - 9\right)$
y-intercept: $\left(- 5\right)$
x-intercepts: $\left(- 5\right) \mathmr{and} \left(+ 1\right)$

Explanation:

To begin, let's convert the given equation
$\textcolor{w h i t e}{\text{XXX}} y = {x}^{2} + 4 x - 5$
into "vertex form": $y = m {\left(x - a\right)}^{2} + b$ (with vertex at $\left(a , b\right)$)

Completing the square:
$\textcolor{w h i t e}{\text{XXX}} y = {x}^{2} + 4 x \textcolor{g r e e n}{+ 4} - 5 \textcolor{g r e e n}{- 4}$

Simplifying and writing as a squared binomial:
$\textcolor{w h i t e}{\text{XXX}} y = {\left(x + 2\right)}^{2} - 9$ (** this is the form I will use for the intercepts later)
or
$\textcolor{w h i t e}{\text{XXX}} y = 1 {\left(x - \left(- 2\right)\right)}^{2} + \left(- 9\right)$
which is in vertex form with vertex at $\left(- 2 , - 9\right)$

The y-intercept is the value of $y$ when $x = 0$
$\textcolor{w h i t e}{\text{XXX}} y = {\left(0 + 2\right)}^{2} - 9 = - 5$

The x-intercepts are the values of $x$ when $y = 0$
$\textcolor{w h i t e}{\text{XXX}} 0 = {\left(x + 2\right)}^{2} - 9$

$\textcolor{w h i t e}{\text{XXX}} \rightarrow {\left(x + 2\right)}^{2} = 9$

$\textcolor{w h i t e}{\text{XXX}} \rightarrow x + 2 = \pm 3$

$\textcolor{w h i t e}{\text{XXX}} \rightarrow x = - 5 \mathmr{and} + 1$

graph{x^2+4x-5 [-12.37, 10.13, -10.305, 0.945]}