# Question #bc867

Dec 13, 2017

Factor it to find the zeros at $x = - 1 , 3$, and find the vertex at $\left(1 , - 8\right)$. Plot those points, and there's your parabola! graph{2x^2-4x-6 [-20, 20, -10, 10]}

#### Explanation:

Factoring is the fastest way to find the zeros of this function.
Factor out the GCF first: $2 {x}^{2} - 4 x - 6 = 2 \left({x}^{2} - 2 x - 3\right)$.
Factor the second part: ${x}^{2} - 2 x - 3 = \left(x - 3\right) \left(x + 1\right)$.
Put it all together: $g \left(x\right) = 2 \left(x - 3\right) \left(x + 1\right)$

The solutions/zeros/roots are where the function crosses the
x-axis, which means y=0 at those points. Set $g \left(x\right) = 0$:
$2 \left(x - 3\right) \left(x + 1\right) = 0$.
There are two ways the product of these three terms equals zero:
If $\left(x - 3\right) = 0$ or if $\left(x + 1\right) = 0$, then $g \left(x\right) = 0$
Solve the two equations to get $x = 3$ and $x = - 1$.
These are the places that the function crosses the x-axis.

Find the vertex:
The x-coordinate of the vertex of a parabola is always at $- \frac{b}{2 a}$. Plug that number back into the original function to get the corresponding y-coordinate.
$- \frac{b}{2 a} = \frac{- \left(- 4\right)}{2 \left(2\right)} = \frac{4}{4} = 1$ is the x-coordinate of the vertex.
Find the y-coordinate: $g \left(1\right) = 2 {\left(1\right)}^{2} - 4 \left(1\right) - 6 = 2 - 4 - 6 = - 8$
The vertex is at $\left(1 , - 8\right)$.
Plot the vertex, plot the zeros, and draw your parabola!