How do you graph the parabola y = x^2 - x - 2 using vertex, intercepts and additional points?

1 Answer
Nov 25, 2017

Firstly, the coefficient of the x^2 term is positive (1) so it must have a minimum and go on infinitely in the positive y direction.

The minimum point (vertex) will be when the gradient is zero, so differentiating y=x^2-x-2 gives doty=2x-1=0 which means that x=0.5 at the minimum. If x=0.5, y=0.5^2-0.5-2=-2.25.
So now that we have the x and y coordinates of the minimum, we know to draw the minimum at (0.5, -2.25).

When x=0, y=0^2-0-2=-2, so the parabola has the point (0,-2), which means it crosses the y-axis at -2.

Factorizing the quadratic gives y=x^2-x-2=(x+1)(x-2), so when y=0, (x+1)(x-2)=0, so either x+1=0, which means x=-1 or x-2=0 which means x=2.

Now we have all the information to draw it. It has a minimum at (0.5, -2.25) crosses the y axis at -2, and crosses the x axis at -1 and 2