# How do you sketch the graph of y=-2(x-1)^2+7 and describe the transformation?

Dec 15, 2017

See method below.

#### Explanation:

Transformation

Let $f \left(x\right) = {x}^{2}$

$f \left(x - 1\right) = {\left(x - 1\right)}^{2}$
This is a translation by 1 to the right.

$2 f \left(x - 1\right) = 2 {\left(x - 1\right)}^{2}$
This is a vertical stretch, scale factor 2.

$- 2 f \left(x - 1\right) = - 2 {\left(x - 1\right)}^{2}$
This is a reflection in the x-axis (y=0).

$- 2 f \left(x - 1\right) + 7 = - 2 {\left(x - 1\right)}^{2} + 7$
This is a translation up by 7.

Put this together; our transformation is:

Reflection in the x-axis, vetical stretch scale factor 2, transformation by $\left(\begin{matrix}1 \\ 7\end{matrix}\right)$.

Sketching

We need to find some points to plot the curve.

$y = - 2 {\left(x - 1\right)}^{2} + 7$

From this completed square form, we can tell that the minimum will be at $\left(1 , 7\right)$

Find the x-intercepts:

Let $y = 0$
$0 = - 2 {\left(x - 1\right)}^{2} + 7$
$- 7 = - 2 {\left(x - 1\right)}^{2}$
$\frac{7}{2} = {\left(x - 1\right)}^{2}$
$\pm \sqrt{\frac{7}{2}} = x - 1$
$x = 1 \pm \sqrt{\frac{7}{2}}$

so $x = 1 + \sqrt{\frac{7}{2}}$ or $x = 1 - \sqrt{\frac{7}{2}}$

Find the y-intercepts:

$y = - 2 \left({x}^{2} - 2 x + 1\right) + 7$
$y = - 2 {x}^{2} + 4 x - 2 + 7$
$y = - 2 {x}^{2} + 4 x + 5$

Let $x = 0 \implies y = 5$
So there is a y-intercept at $\left(0 , 5\right)$

The co-efficient of ${x}^{2}$ is negative, so it is an upside-down-U-shaped curve.

The graph should look like this:

graph{-2(x-1)^2+7 [-14.16, 17.88, -6.48, 9.54]}