Solve 0≤x2−8x≤20 algebraically. Graph the solution.

Answer 1

For y = x^2 - 8x = x^2 - 8x + 16 - 16 = (x-4)^2 - 16

This is a parabola facing up with vertex (4, -16)

Thus, as 0 > -16 and the parabola faces upwards, 0 <= x^2 - 8x <= 20 will have a solution that includes two intervals.

If 0 = x^2 - 8x = x(x -8), x has two solutions, 0 and 8 (note how they are equidistant from 4).

If 20 = x^2 - 8x, 0 = x^2 - 8x - 20 = (x - 10)(x +2), x has two solutions, -2 and 10 (again, notice that these are equidistant from 4).

Thus, our solution to 0 <= x^2 - 8x <= 20 is, from our values calculated above, [-2, 0] ∪ [8, 10]

To graph this, while one could argue that the graph is a number line with these two x intervals indicated, what would make more sense would be to graph y =x^2 - 8x over the x interval [-3, 11]

You might then shade the region 0<= y <= 20 as well as -2 <= x<=0 and 8 <= x <= 10 (a horizontal stripe and 2 vertical stripes) to show the solution.