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2 years ago

9

Which equation is represented by the graph below? On a coordinate plane, a curve starts at (0, negative 3) in quadrant 4 and the

n increases into quadrant 1 and approaches y = 3. y = e Superscript x y = e Superscript x Baseline minus 1 y = l n x y = l n x minus 1

1 answer:

ohaa [14]2 years ago

4 0

Answer:

Its C

Step-by-step explanation:

My friend who is very smart told me

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2 years ago

I’m Really lost if I could get a answer it would be greatly appreciated

Nataly [62]

<h3>Answers:</h3>

$f(g(x)) = \sqrt{x^2+5}+5\\\\g(f(x)) = x+30+10\sqrt{x-1}$

================================================

Work Shown:

Part 1

$f(x) = \sqrt{x-1}+5\\\\f(g(x)) = \sqrt{g(x)-1}+5\\\\f(g(x)) = \sqrt{x^2+6-1}+5\\\\f(g(x)) = \sqrt{x^2+5}+5\\\\$

Notice how I replaced every x with g(x) in step 2. Then I plugged in g(x) = x^2+6 and simplified.

------------------

Part 2

$g(x) = x^2+6\\\\g(f(x)) = \left(f(x)\right)^2+6\\\\g(f(x)) = \left(\sqrt{x-1}+5\right)^2+6\\\\g(f(x)) = \left(\sqrt{x-1}\right)^2+2*5*\sqrt{x-1}+\left(5\right)^2+6\\\\g(f(x)) = x-1+10\sqrt{x-1}+25+6\\\\g(f(x)) = x+30+10\sqrt{x-1}\\\\$

In step 4, I used the rule (a+b)^2 = a^2+2ab+b^2

In this case, a = sqrt(x-1) and b = 5.

You could also use the box method as a visual way to expand out $\left(\sqrt{x-1}+5\right)^2$

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2 years ago

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Gnoma [55]

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