# How do you solve the absolute value equation y=-2abs(5x+8)+4 and find the vertex, x intercepts and y intercept?

May 28, 2015

Given $y = - 2 \left\mid 5 x + 8 \right\mid + 4$
If $\left(5 x + 8\right) \le 0$ then $y = - 2 \left(- 5 x - 8\right) + 4$ which is a linear equation.
Similarly
If $\left(5 x + 8\right) \ge 0$ then $y = - 2 \left(5 x + 8\right) + 4$ which is also a linear equation

Any vertex must exist at the point where these two condition meet;
That is when $\left(5 x + 8\right) = 0$
at $\left(x , y\right) = \left(- \frac{8}{5} , 4\right)$

The y-intercept occurs when $x = 0$
$y = - 2 \left\mid 5 \left(0\right) + 8 \right\mid + 4 = - 12$

The x-intercepts occur when $y = 0$
Case 1: $\left(5 x + 8\right) \ge 0$
$0 = - 2 \left(5 x + 8\right) + 4$
$5 x + 8 = 2$
$x = - \frac{6}{5}$

Case 2: $\left(5 x + 8\right) < 0$
0 = -2(-5x-8)+4 (-5x-8) = 2 -5x = 10 x = -2#

In summary

the critical point is at $\left(- \frac{8}{5} , 4\right)$
the y-intercept is at $\left(- 12\right)$
the x-intercepts are at $\left(- \frac{6}{5}\right)$ and
graph{-2*abs(5x+8)+4 [-5.696, 4.17, -0.437, 4.493]}