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

Mathematics

1 answer:

sineoko [7]4 years ago

7 0

Answer:

Option C. −4≤t≤−3

Step-by-step explanation:

we have

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

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

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