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

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 $\dot{y} = 2 x - 1 = 0$ 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 $y = {x}^{2} - x - 2 = \left(x + 1\right) \left(x - 2\right)$, so when $y = 0$, $\left(x + 1\right) \left(x - 2\right) = 0$, 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