# How do you solve -3/4x^2+4x-8<0 by graphing?

Nov 5, 2017

Graph $y = - \frac{3}{4} {x}^{2} + 4 x - 8$ with the vertex, y-intercept and (if any) x-intercepts.

#### Explanation:

Let $y = - \frac{3}{4} {x}^{2} + 4 x - 8$.
[Step1] First, you need to complete the square.
$y = - \frac{3}{4} {x}^{2} + 4 x - 8$
$y = - \frac{3}{4} \left({x}^{2} - \frac{16}{3} x\right) - 8$
$y = - \frac{3}{4} \left\{{\left(x - \frac{8}{3}\right)}^{2} - {\left(\frac{8}{3}\right)}^{2}\right\} - 8$
$y = - \frac{3}{4} {\left(x - \frac{8}{3}\right)}^{2} + \frac{3}{4} \cdot \frac{64}{9} - 8$
$y = - \frac{3}{4} {\left(x - \frac{8}{3}\right)}^{2} - \frac{8}{3}$

Its vertex is $\left(\frac{8}{3} , - \frac{8}{3}\right)$ and the graph is convex upward.

[Step2] Determine the intercepts.
Put $x = 0$ to the function and $y$-intercept is $- 8$.
However, there is no $x$-interscepts, since $y$ is always
negative.

graph{-3/4(x-8/3)^2-8/3 [-1, 6, -10, 2]}

Therefore, the domain for x that satisfies the given inequation is $- \infty < x < \infty$, or, any real number.