# How do you graph the parabola y= 2x^2 – 4x – 6 using vertex, intercepts and additional points?

Jul 28, 2018

$y = 2 {x}^{2} - 4 x - 6$ is a quadratic equation in standard form:

$y = a {x}^{2} + b x + c$,

where:

$a = 2$, $b = - 4$, $c = - 6$

Vertex: maximum or minimum point of a parabola

Since $a > 0$, the vertex is the minimum point and the parabola opens upward.

To find the x-coordinate of the vertex , use the formula for the axis of symmetry:

$x = \frac{- b}{2 a}$

$x = \frac{- \left(- 4\right)}{2 \cdot 2}$

$x = \frac{4}{4}$

$x = 1$

To find the y-coordinate of the vertex , substitute $1$ for $x$ in the equation and solve for $y$.

$y = 2 {\left(1\right)}^{2} - 4 \left(1\right) - 6$

$y = 2 - 4 - 6$

$y = - 8$

The vertex is $\left(1 , - 8\right)$. Plot this point.

Y-intercept: the value of $y$ when $x = 0$

Substitute $0$ for $x$ and solve for $y$.

$y = 2 {\left(0\right)}^{2} - 4 \left(0\right) - 6$

$y = - 6$

The y-intercept is $\left(0 , - 6\right)$. Plot this point.

X-intercepts: values of $x$ when $y = 0$

Substitute $0$ for $y$ and solve for $x$.

$2 {x}^{2} - 4 x - 6 = 0$ $\leftarrow$ I switched sides to get the variable on the left-hand side.

Factor out the common factor $2$.

$2 \left({x}^{2} - 2 x - 3\right) = 0$

Find two numbers that when multiplied equal $- 3$ and when added equal $- 2$. The numbers $1$ and $- 3$ meet the requirement.

$2 \left(x + 1\right) \left(x - 3\right) = 0$

Solve each binomial for $0$.

$x + 1 = 0$

$x = - 1$

$x - 3 = 0$

$x = 3$

The x-intercepts are $\left(- 1 , 0\right)$ and $\left(3 , 0\right)$. Plot the points.

Additional points: choose values for $x$ and solve for $y$.

$x = 2$

$y = 2 {\left(2\right)}^{2} - 4 \left(2\right) - 6$

$y = 8 - 8 - 6$

$y = - 6$

Additional point 1 is $\left(2 , - 6\right)$. Plot this point.

$x = - 2$

$y = 2 {\left(- 2\right)}^{2} - 4 \left(- 2\right) - 6$

$y = 8 + 8 - 6$

$y = 10$

Additional point 2 is $\left(- 2 , 10\right)$. Plot this point.

$x = 4$

$y = 2 {\left(4\right)}^{2} - 4 \left(4\right) - 6$

$y = 32 - 16 - 6$

$y = 10$

Additional point 3 is $\left(4 , 10\right)$. Plot this point.

Plot the points and sketch a parabola through them. Do not connect the dots.

graph{y=2x^2-4x-6 [-12.21, 10.29, -8.505, 2.745]}