Question

How do you solve this quadratic inequality: y≤4x²+4x+4?

I need to find the two x's and graph

Answer 1

Graph the parabola, so start by completing the square.

`y = 4(x^2 + x) + 4`

`y = 4(x^2 + x+ 1/4 - 1/4) + 4`

`y = 4(x^2 +x + 1/4) - 1 + 4`

`y = 4(x + 1/2)^2 + 3`

So the vertex is at `(-1/2, 3)`. The parabola opens up because `a` in `y = a(x- p)^2 + q` is positive. There is a y-intercept at `x= 4`.

When you draw the parabola, make sure to use a solid line, because the symbol is ≤ and not < (in the latter case you would use a dotted line.

You should get the following:

graph{y = 4x^2 + 4x + 4 [-20, 20, -10, 10]}

Now select a test point on either side. Let it be `(0, 1)`.

`1 ≤ 4(0)^2 + 4(0) + 4`

`1 ≤ 4`

This is true, therefore, you should shade the area outside the parabola.

Hopefully this helps!