Quadratic Inequalities question?

y = x^2 - 2x is a parabola opening upward with vertex at (1,-1). This can be seen by completing the square and rewriting the equation as y = (x - 1)^2 - 1. Graphing the parabola will help you visualize the x values that you are looking for. To find what values of x make y ≤ 8, we find out what values of x make y = 8.

x^2 - 2x = 8

x^2 - 2x - 8 = 0

(x - 4)(x + 2) = 0

x = 4, -2

By looking at the graph of y = x^2 - 2x, we see that the values of x for which y ≤ 8 are

Answer: -2 ≤ x ≤ 4

Add 8 to both sides to get x^2-2x-8=0. This factors to

(x-4)(x+2)=0 leading to x-intercepts of 4 and -2.

Now find the vertex. The x-coordinate is the average of the x-intercepts (1). The y-coordinate you get by plugging 1 into the x^2-2x-8, which is -9. So, the vertex is (-1,-9). Sketch the parabola.

You can see where the parabola is below the x-axis, which is between -2 and 4. So the solution becomes -2 ≤x≤4.

You're exactly right; you need to find where it crosses the x-axis. This amounts to finding the roots of the equation. Subtract 8 from both sides to make factoring easier.

Now we have x^2 - 2x - 8 <= 0

Factor into (x -4)(x+2).

So the graph crosses the x-axis at 4 and -2. Do you know how to take it from here?

x^2 - 2x ≤ 8

x^2 - 2x - 8 ≤ 0 [subtracted 8 from both sides]

(x-4)(x+2) ≤ 0 [factored]

With the zero-product property you can then put it into:

x-4 ≤ 0

x+2 ≤ 0

x ≤ 4 [add 4 to both sides]

x ≤ -2 [subtract 2 from both sides]

Answer:

x ≤ 4, x ≤ -2

OR

-2 ≤ x ≤ 4