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H(t)=(t+3)^2+5

Over which interval does h have a negative average rate of change?

Choose 1 answer:

(Choice A)

−2≤t≤0

(Choice B)

1≤t≤4

(Choice C)

−4≤t≤−3

(Choice D)

−3≤t≤4

Respuesta :

Answer:

Option C. −4≤t≤−3

Step-by-step explanation:

we have

[tex]H(t)=(t+3)^2+5[/tex]

This is the equation of a vertical parabola written in vertex form

The parabola open upward (the leading coefficient is positive)

The vertex is a minimum

The vertex is the point (-3,5)

The function is increasing in the interval [-3,∞)

The function is decreasing in the interval (-∞,-3]

The function will have a negative average rate when the function will be decreasing

therefore

the answer is option C