# Question #be5ca

Dec 4, 2017

The first one should look like this: graph{-2(x+2)^2+6 [-10, 10, -5, 5]}
The second one should look like this: graph{1/2(x)^2-6x+13 [-10, 10, -5, 5]}

#### Explanation:

Our method for graph the parabola will be like this:
We find the values of a, h, and k.
Then, we graph the other four points by doing the following:
We will do h-2a, h-a, h+a, and h+2a for the x values and plug them in the equation to find the y values.
The minimum/maximum point of a parabola is (h,k)

$- 2 {\left(x + 2\right)}^{2} + 6$:

This equation is in the form $a {\left(x - h\right)}^{2} + k$

Therefore,
$a = - 2$
$h = - 2$
$k = 6$

The x values are:
$- 2 - 2 \left(- 2\right) = 2$
$- 2 - \left(- 2\right) = 0$
$- 2$
$- 2 + \left(- 2\right) = - 4$
$- 2 + 2 \left(- 2\right) = - 6$

Plug in the values in the equation to get the following y values.
$- 2 - 2 \left(- 2\right) = 2$ =>-28
$- 2 - \left(- 2\right) = 0$ =>-2
$- 2$ =>6
$- 2 + \left(- 2\right) = - 4$ =>-2
$- 2 + 2 \left(- 2\right) = - 6$ =>-28

Plot the points an you have the answer!

$\frac{1}{2} {\left(x\right)}^{2} - 6 x + 13$:

This equation is in the form $a {x}^{2} + b x + c$
We have to know that $h = \frac{- b}{2 a}$
Plug in this value to the equation to get k.
Therefore,
$a = \frac{1}{2}$
$h = 6$
$k = - 5$

The x values are:
$6 - 2 \left(\frac{1}{2}\right) = 5$
$6 - \frac{1}{2} = 5 \frac{1}{2}$
$6$
$6 + \frac{1}{2} = - 6 \frac{1}{2}$
$6 + 2 \left(\frac{1}{2}\right) = 7$

Plug in the values in the equation to get the following y values.
$6 - 2 \left(\frac{1}{2}\right) = 5$ =>-4.5
$6 - \frac{1}{2} = 5 \frac{1}{2}$ =>-4.875
$6$ =>-5
$6 + \frac{1}{2} = - 6 \frac{1}{2}$=>-4.875
$6 + 2 \left(\frac{1}{2}\right) = 7$=>-4.5

Plot the points an you have the answer!