Question

How do you graph `y>4/3x^2-12x+29`?

Answer 1

First, graph the function
`y = (4x^2)/3 - 12x + 29.`
The area that is above the parabola graph is the solution set.
graph{(4x^2)/3 - 12x + 29 [-20, 20, -10, 10]}