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

Apr 4, 2018

y-intercept
$y = - 5$

x-intercepts
${x}_{1} = - 5$
${x}_{2} = 1$

#### Explanation:

an x-intercept is a point where $y = 0$, and
a y-intercept is a point where $x = 0$

x-intercepts: $0 = {x}^{2} + 4 x - 5$

${x}^{2} + 4 x - 5 + {\left(\frac{4}{2}\right)}^{2} = 0 + {\left(\frac{4}{2}\right)}^{2}$ complete the square
${\left(x + 2\right)}^{2} - 5 = 4$
${\left(x + 2\right)}^{2} = 9$
x + 2=± 3
x =-2± 3

${x}_{1} = - 5$
${x}_{2} = 1$

y-intercept: $y = {\left(0\right)}^{2} + 4 \left(0\right) - 5 = - 5$

Check the graph

graph{x^2 + 4x - 5 [-9.96, 10.04, -8.72, 1.28]}