# What are the intercepts of y = 4(x - 5)+x^2?

Apr 16, 2017

The x-intercepts are the point of $\left(2.899 , 0\right)$ and $\left(- 6.899 , 0\right)$, the y-intercept is $\left(0 , - 20\right)$

#### Explanation:

For y-intercept(s), let $x = 0$

Doing so,

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

$y = 4 \left(- 5\right)$

$y = - 20$

Therefore the y-intercept is $\left(0 , - 20\right)$

For x-intercept(s), let $y = 0$

Doing so,

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

${x}^{2} + 4 x - 20 = 0$

Use the quadratic formula (i'll let you do that),

${x}_{1} = - 6.899$ and ${x}_{2} = 2.899$

Therefore the x-intercepts are the point of $\left(2.899 , 0\right)$ and $\left(- 6.899 , 0\right)$, the y-intercept is $\left(0 , - 20\right)$