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kaheart [24]
3 years ago
12

Which of the following describes the transformation of g(x) = -(2)x+4 -2 from the parent function f(x)= 2^x

Mathematics
2 answers:
Dmitrij [34]3 years ago
7 0

Answer:

Shift 4 units left, reflected over x-axis, shift two units down.

Step-by-step explanation:

We have been given formula of a parent function f(x)=2^x and we are asked to find the transformations that are used to get g(x)=-(2)^{x+4}-2 from our given parent function.

Let us recall transformation rules.

f(x-a)\rightarrow \text{Graph shifted to right by a units},

f(x+a)\rightarrow \text{Graph shifted to left by a units},

f(x)+a\rightarrow \text{Graph shifted upwards by a units},

f(x)-a\rightarrow \text{Graph shifted downwards by a units},

f(-x)\rightarrow \text{Graph reflected across y-axis},

-f(x)\rightarrow \text{Graph reflected across x-axis}.

Upon looking at our given f(x) and g(x),we can see these transformations:

  • Our parent function is shifted to left by 4 units.
  • Our parent function is reflected over x-axis.
  • Our parent is shifted downwards by 2 units.

Therefore, option A is the correct choice.

Nataly [62]3 years ago
4 0
Ok, so what happend was 4 was added to every x, the whole thing was multiplied by -1 and -2 was added to the whole function

so
adding 4 to every x shifts the graph to the left 4 units
multiplying the whole function by -1 reflects it over the x axis
adding -2 to the whole function moves it down2 units

so that would be the first option
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Find the area of each figure. Round to the nearest tenth. Use 3.14 or 22/7
sertanlavr [38]

Answer:

d. 28.3

Step-by-step explanation:

Given

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3 0
3 years ago
john is 15 years old. john's mom takes him and his younger soster to a football game. tickets are $22 for adults and $16 for chi
Maksim231197 [3]

Answer:

$38

Step-by-step explanation:

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cost of ticket for youngster = $16

______________________________

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5 0
3 years ago
Inverse laplace of [(1/s^2)-(48/s^5)]
Katen [24]
**Refresh page if you see [ tex ]**

I am not familiar with Laplace transforms, so my explanation probably won't help, but given that for two Laplace transform F(s) and G(s), then \mathcal{L}^{-1}\{aF(s)+bG(s)\} = a\mathcal{L}^{-1}\{F(s)\}+b\mathcal{L}^{-1}\{G(s)\}

Given that \dfrac{1}{s^2} = \dfrac{1!}{s^2} and -\dfrac{48}{s^5} = -2\cdot\dfrac{4!}{s^5}

So you have \mathcal{L}^{-1}\left\{\dfrac{1}{s^2} - 2\cdot\dfrac{4!}{s^5}\right\} = \mathcal{L}^{-1}\left\{\dfrac{1}{s^2}\right\} - 2\mathcal{L}^{-1}\left\{\dfrac{4!}{s^5}\right\}

From Table of Laplace Transform, you have \mathcal{L}\{t^n\} = \dfrac{n!}{s^{n+1}} and hence \mathcal{L}^{-1}\left\{\dfrac{n!}{s^{n+1}}\right\} = t^n

So you have \mathcal{L}^{-1}\left\{\dfrac{1}{s^2}\right\} - 2\mathcal{L}^{-1}\left\{\dfrac{4!}{s^5}\right\} = \boxed{t-2t^4}.

Hope this helps...
7 0
3 years ago
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