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Nookie1986 [14]
3 years ago
12

How is the graph of the parent function, y = StartRoot x EndRoot transformed to produce the graph of y = StartRoot negative 2 x

EndRoot? It is translated horizontally by 2 units and reflected over the x-axis. It is translated horizontally by 2 units and reflected over the y-axis. It is horizontally compressed by a factor of 2 and reflected over the x-axis. It is horizontally compressed by a factor of 2 and reflected over the y-axis.

Mathematics
2 answers:
tino4ka555 [31]3 years ago
7 0

The graph of the parent function is horizontally compressed by a factor of 2 and reflected over the y-axis ⇒ Last answer

Step-by-step explanation:

Let us revise some transformation

A horizontal compression (or shrinking) is the squeezing of the graph toward the y-axis.  

A horizontal stretching is the stretching of the graph away from the y-axis  

  • If k > 1, the graph of y = f(k•x) is the graph of f(x) horizontally shrunk (or compressed) by dividing each of its x-coordinates by k.
  • If 0 < k < 1 (a fraction), the graph of y = f(k•x) is the graph of f(x) horizontally stretched by dividing each of its x-coordinates by k
  • If k should be negative, the horizontal stretch or shrink is followed by a reflection across the y-axis

∵  y = √x is a parent function

∵ Its graph transformed to produce the graph of y = √(-2x)

- That means x is multiplied by a factor k, then it is compressed

   or stretched horizontally

∵ k = -2

- The factor is -2, where 2 is greater than 1, then it is compressed

   horizontally as the first rule above and the negative means

   reflected across the y-axis as the third rule above

∴ The graph of the parent function is compressed horizontally and

    reflected across the y-axis

The graph of the parent function is horizontally compressed by a factor of 2 and reflected over the y-axis

Look to the attached figure for more understand

Learn more:

You can learn more about transformation in brainly.com/question/9381523

#LearnwithBrainly

andrew-mc [135]3 years ago
7 0

Answer:

D on edge

Step-by-step explanation:

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\begin{cases}a^2+b^2-2b+1=r^2\\a^2+b^2+4a+2b+5=r^2\end{cases}\\--------------(-)\\\\a^2+b^2-2b+1-a^2-b^2-4a-2b-5=r^2-r^2\\\\-4a-4b-4=0\qquad|:(-4)\\\\\boxed{-a-b-1=0}

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\begin{cases}a^2+b^2+2a-4b+5=r^2\\a^2+b^2-2b+1=r^2\end{cases}\\--------------(-)\\\\a^2+b^2+2a-4b+5-a^2-b^2+2b-1=r^2-r^2\\\\2a-2b+4=0\qquad|:2\\\\\boxed{a-b+2=0}

Now we can easly calculate a and b:

\begin{cases}-a-b-1=0\\a-b+2=0\end{cases}\\--------(+)\\\\-a-b-1+a-b+2=0+0\\\\-2b+1=0\\\\-2b=-1\qquad|:(-2)\\\\\boxed{b=\frac{1}{2}}\\\\\\\\a-b+2=0\\\\\\a-\dfrac{1}{2}+2=0\\\\\\a+\dfrac{3}{2}=0\\\\\\\boxed{a=-\frac{3}{2}}

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a^2+b^2-2b+1=r^2\\\\\\\left(-\dfrac{3}{2}\right)^2+\left(\dfrac{1}{2}\right)^2-2\cdot\dfrac{1}{2}+1=r^2\\\\\\\dfrac{9}{4}+\dfrac{1}{4}-1+1=r^2\\\\\\\dfrac{10}{4}=r^2\\\\\\\boxed{r^2=\frac{5}{2}}

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