What is the vertex if the function f(x)=-x^2-2x-1

Mathematics

1 answer:

Mila [183]3 years ago

7 0

Vertex=(-1,0)

x^2-2x-1

y=− ( x + 1 )^2 + 0

Use the vertex form, y=a(x−h)^2+k, to determine the values of a, h, and k.

Find the vertex (h,k) = (-1,0)

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AlekseyPX

I- huh I don’t understand the question

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3 years ago

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jek_recluse [69]

<h3>Answer: Choice D</h3>

$(f \circ g)(x) = x+12$

Domain = [-3, infinity)

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Work Shown:

$f(x) = x^2 + 9\\\\f(g(x)) = (g(x))^2 + 9\\\\f(g(x)) = (\sqrt{x+3})^2 + 9\\\\f(g(x)) = x+3 + 9\\\\f(g(x)) = x+12\\\\(f \circ g)(x) = x+12\\\\$

In step 2, we replaced every x with g(x)

In step 3, we plugged in g(x) = sqrt(x+3)

The domain of g(x) is [-3, infinity), so this is the domain of $(f \circ g)(x)$ as well since the composite function depends entirely on g(x). Put another way: the input of f(x) depends on the output of g(x), so that's why the domains match up.