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LiRa [457]
3 years ago
6

For problems 1 through 3, please include –

Mathematics
1 answer:
Roman55 [17]3 years ago
5 0

Answer:

Step-by-step explanation:

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The function g(x)=x^2+3. The f(x)=g(x+2)
vekshin1

Answer:

f(x)=x^2+6x+12

Step-by-step explanation:

Given:

The function 'g(x)' is given as:

g(x)=x^2+3

Now, the function 'f(x)' is given as:

f(x)=g(x+2)

So, the function 'f(x)' is a transformation of 'g(x)'.

In order to find f(x), we replace 'x' by (x + 2) in the 'g(x)' function equation. This gives,

g(x+2)=(x+2)^2+3

Using the identity (a+b)^2=a^2+b^2+2ab, we get:

g(x+2)=x^2+3^2+6x+3\\\\g(x+2)=x^2+9+6x+3\\\\g(x+2)=x^2+6x+12

Hence the function f(x) is given as:

f(x)=x^2+6x+12

4 0
4 years ago
Which ordered pairs match the table?
ArbitrLikvidat [17]
The answer is a. (0,2) , (1,2), (2,3), (1,4) , (3,1)
this is because (x, y) . x is first then y
6 0
3 years ago
Read 2 more answers
Determine the domain and range of (g ○ f)(x) if f of x is equal to 4 over the quantity x squared minus 4 end quantity and g(x) =
lara [203]

The domain and range of a function are the possible <em>x and y values </em>of the function.

<em>The domain and the range of the function is: (a) </em>\mathbf{D:\{x \in R|x \ne -2,2\}}<em> and </em>\mathbf{R:\{(-\infty,1) \cup (2,\infty)\}}<em />

The functions are given as:

\mathbf{f(x) = \frac{4}{x^2 - 4}}

\mathbf{g(x) = x + 2}

(g o f)(x) is calculated as:

\mathbf{(g\ o\ f)(x) = g(f(x))}

So, we have:

\mathbf{(g\ o\ f)(x) = \frac{4}{x^2 - 4} + 2}

Take LCM

\mathbf{(g\ o\ f)(x) = \frac{4 + 2x^2 - 8}{x^2 - 4} }

\mathbf{(g\ o\ f)(x) = \frac{2x^2 - 4}{x^2 - 4} }

Represent the denominator as follows, to calculate the domain

\mathbf{x^2 - 4 \ne 0 }

Add 4 to both sides

\mathbf{x^2 \ne 4 }

Take square roots of bot sides

\mathbf{x \ne \±2 }

Hence, the domain of the function is:

\mathbf{D:\{x \in R|x \ne -2,2\}}

On the graph of \mathbf{(g\ o\ f)(x) = \frac{2x^2 - 4}{x^2 - 4} } (see attachment), the function does not have a value from <em>y = 1 to 2.</em>

Hence, the range is:

\mathbf{R:\{(-\infty,1) \cup (2,\infty)\}}

Read more about domain and range at:

brainly.com/question/1632425

5 0
2 years ago
Read 2 more answers
<img src="https://tex.z-dn.net/?f=%20%7B3%7D%5E%7Bx%20%5E%7B2%20-%201%7D%20%7D%20%20%3D%20%20%5Cfrac%7B%20%7B27%7D%5E%7B%20-%20x
vitfil [10]

Answer:

-1/4

Step-by-step explanation:

4 0
3 years ago
What is the center of a circle represented by the equation (x-5)2+(y+6)2=42?
Ilia_Sergeevich [38]

\bf \textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \qquad center~~(\stackrel{}{ h},\stackrel{}{ k})\qquad \qquad radius=\stackrel{}{ r} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ (x-5)^2+(y+6)^2=42\implies [x-\stackrel{h}{5}]^2+[y-(\stackrel{k}{-6})]^2=(\stackrel{r}{\sqrt{42}})^2~~ \begin{cases} \stackrel{center}{(5,-6)}\\\\ \stackrel{radius}{\sqrt{42}} \end{cases}

4 0
4 years ago
Read 2 more answers
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