# How do you use the important points to sketch the graph of y = 2x^2 + x − 15?

Feb 10, 2018

See explanation.

#### Explanation:

We can factor $2 {x}^{2} + x - 15 = \left(2 x - 5\right) \left(x + 3\right)$.

From this we know the $x$-intercepts are $\left(\frac{5}{2} , 0\right)$ and $\left(- 3 , 0\right)$.

The $x$-coordinate of the vertex is located at the average of the $x$-coordinates of the intercepts, so at $x = - \frac{1}{4}$. The $y$-value of the vertex can be found by substitution:

$\left(2 \left(- \frac{1}{4}\right) - 5\right) \left(- \frac{1}{4} + 3\right) = - \frac{121}{8}$.

So the vertex is $\left(- \frac{1}{4} , - \frac{121}{8}\right)$.

The $y$-intercept is found by substituting $x = 0$ to get $y = - 15$, so the $y$-intercept is $\left(0 , - 15\right)$.

Plotting all of these points and sketching a parabola through them we get the graph we want.

graph{y=2x^2+x-15 [-16.83, 23.17, -14.96, 5.04]}