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

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Answer 1 · 2017-12-15T18:37:20

See method below.

Transformation

Let $f(x)=x^2$

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

$2f(x-1)=2(x-1)^2$
This is a vertical stretch, scale factor 2.

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

$-2f(x-1)+7=-2(x-1)^2+7$
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 $((1),(7))$ .

Sketching

We need to find some points to plot the curve.

$y=-2(x-1)^2+7$

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

Find the x-intercepts:

Let $y=0$
$0=-2(x-1)^2+7$
$-7=-2(x-1)^2$
$7/2=(x-1)^2$
$+-sqrt(7/2)=x-1$
$x=1+-sqrt(7/2)$

so $x=1+sqrt(7/2)$ or $x=1-sqrt(7/2)$

Find the y-intercepts:

$y=-2(x^2-2x+1)+7$
$y=-2x^2+4x-2+7$
$y=-2x^2+4x+5$

Let $x=0 => y=5$
So there is a y-intercept at $(0, 5)$

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]}

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