# How do you identify the important parts of y = x^2 - x - 2 to graph it?

Sep 25, 2015

The axis of symmetry is $x = \frac{1}{2}$.
The vertex is $\left(\frac{1}{2} , - 2 \frac{1}{4}\right)$.

#### Explanation:

$y = {x}^{2} - x - 2$ is a quadratic equation of the form $a {x}^{2} + b x + c$, where $a = 1 , b = - 1 , \mathmr{and} c = - 2$.

Axis of Symmetry

The graph of a quadratic equation is a parabola. First find the axis of symmetry. This is the vertical line that divides the parabola in half. The formula for the axis of symmetry is $x = \frac{- \left(b\right)}{2 a}$.

$x = \frac{- \left(b\right)}{2 a} = \frac{- \left(- 1\right)}{2 \cdot 1} =$

The axis of symmetry is $x = \frac{1}{2}$.

Vertex

Now find the vertex, which is the maximum or minimum point on the parabola. In this case it will be the minimum point. $x = \frac{1}{2}$ is the $x$ value for the vertex. To find the $y$ value, substitute $\frac{1}{2}$ for $x$ in the equation and solve for $x$.

$y = {\left(\frac{1}{2}\right)}^{2} - \frac{1}{2} - 2 =$

$y = \frac{1}{4} - \frac{1}{2} - 2 =$

$y = \frac{1}{4} - \frac{2}{4} - \frac{8}{4} =$

$y = - 2 \frac{1}{4}$

The vertex is $\left(\frac{1}{2} , - 2 \frac{1}{4}\right)$.

Next determine some points by substituting values for $x$ on both sides of the axis of symmetry. Plot the vertex and the other points. Sketch a parabola through the points. Do not connect the dots.

$x = - 1 ,$ $y = 0$
$x = 0 ,$ $= - 2$
$x = \frac{1}{2} ,$ $y = - 2 \frac{1}{4}$
$x = 1 ,$ $y = - 2$
$x = 2 ,$ $y = 0$

graph{y=x^2-x-2 [-10, 10, -5, 5]}