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The graph of $ y = \sqrt{3x - x^2} $ is given. Use transformations to create a function whose graph is as shown.

The graph of $y=f(x)=\sqrt{3 x-x^{2}}$ has been shifted 4 units to the left, reflected about the $x$ -axis, and shifted downward 1 unit. Thus, a function describing the graph is $\begin{array}{ccc}y= & \underbrace{-1}_{\text {reflect }} & f(x+4) & -1 \\ & \text { about } x \text { -axis } & 4 \text { units left } & \text { I unit left }\end{array}$

This function can be written as $$\begin{aligned}

y &=-f(x+4)-1=-\sqrt{3(x+4)-(x+4)^{2}}-1 \\

&=-\sqrt{3 x+12-\left(x^{2}+8 x+16\right)}-1=-\sqrt{-x^{2}-5 x-4}-1

\end{aligned}$$

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Jp R.

September 10, 2019

You forgot about the reflection across the y axis as well.

Shannan G.

April 7, 2021

Sketch the graph of???? = ?2 ? ???? graph starting with graph of????? by utilizing transformation. Utilize this graph to build a picture of derivative of this function. Use the definition of a derivative to find f’ . What are the domains of f and f’

Johns Hopkins University

Oregon State University

Harvey Mudd College

Baylor University

All right, let's find the equation of this graph, which is a transformation of the first graph. So the order of the steps, I think, does make a difference. So this is the order. I would put the steps in. It looks like the first thing that happened was the graph shifted to the left four. And that would put it right about here roughly. And then it reflected across the X axis, which moved it to here. And then the whole thing shifted down, which moved it to its current position. So shifting it to the left four, that would require adding for two X. So we have the equation. Why equals the square root of three times a quantity X plus four minus the quantity X plus four squared and then reflecting it across the X axis. That requires multiplying the function by a negative one. So I get why equals the opposite of what we just had and then finally shifting the whole thing down. One requires us to to subtract one from the entire equation. So outside the radical after all of that, we have minus one. So there's our equation. Now, if we want to simplify it. We could do some work on the inside of the radical, So what we could do is distribute the three. So we get three X Plus 12 and we could multiply the binomial X plus four Times X plus four. And that's going to give us X squared plus eight X plus 16 and then we can distribute the negative outside of the parentheses. So inside the radical we have three X plus 12 minus X squared, minus eight X minus 16 and then we'll combine the like terms. So now inside the radical we have the opposite of X squared minus five X minus four, and we still have the minus one on the outside, and we still have the multiplied by negative one on the outside as well.