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

1 Answer
Jacob
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

Related questions
See all questions in Quadratic Functions and Their Graphs
Impact of this question
1316 views around the world
You can reuse this answer
Creative Commons License