# How do you find the vertex and the intercepts for f(x) = x^2 + 4x - 5?

Nov 9, 2017

$\text{see explanation}$

#### Explanation:

"given a parabola in standard form ";y=ax^2+bx+c

$\text{then the x-coordinate of the vertex can be found using}$

•color(white)(x)x_(color(red)"vertex")=-b/(2a)

$f \left(x\right) = {x}^{2} + 4 x - 5 \text{ is in standard form}$

$\text{with } a = 1 , b = 4 , c = - 5$

$\Rightarrow {x}_{\textcolor{red}{\text{vertex}}} = - \frac{4}{2} = - 2$

$\text{substitute this value into "f(x)" for y coordinate}$

${y}_{\textcolor{red}{\text{vertex}}} = {\left(- 2\right)}^{2} + 4 \left(- 2\right) - 5 = - 9$

$\Rightarrow \textcolor{m a \ge n t a}{\text{vertex }} = \left(- 2 , - 9\right)$

$\text{to obtain the intercepts}$

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

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

$x = 0 \to y = - 5 \leftarrow \textcolor{red}{\text{y-intercept}}$

$y = 0 \to {x}^{2} + 4 x - 5 = 0$

$\text{the factors of - 5 which sum to + 4 are + 5 and - 1}$

$\Rightarrow \left(x + 5\right) \left(x - 1\right) = 0$

$\text{equate each factor to zero and solve for x}$

$x + 5 = 0 \Rightarrow x = - 5$

$x - 1 = 0 \Rightarrow x = 1$

$x = - 1 \text{ and "x=-5larrcolor(red)"x-intercepts}$
graph{x^2+4x-5 [-20, 20, -10, 10]}