A function and an interval of its independent variable are given. The endpoints

Question

A function and an interval of its independent variable are given. The endpoints of the interval are associated with the points P and Q on the graph of the After t seconds, an object dropped from rest falls a distance d-36F, where d is measured in feet and 2 s ts 5 a. Sketch a graph of the function and the secant line through P and Q parts a and b. b. Find the siope of the secant line in part (a), and interpret your answer in terms of an average rate of change over the interval. Include units in your answer a. OA. B. 100) 21 21 2 46 8 10 10 0 246 8 10 0 246 8 10 b. The slope of the secant line is . This means the object falls at an average 1 of | over the interval 2 s ts 5. Click to select your answer(s).

Explanation / Answer

a> The distance function is d(t) = 36t2 and t E [2 , 5]

When t = 2 , d(2) = 36(2)2 = 144

When t = 5 , d(5) = 36(5)2 = 900

So the two end-points on the graph are (2 , 144) and (5 , 900)

From the four graphs given to use :

Graph B : at t= 2 , the value of d is between d = 2880 and d = 3600 . But our start point is (2 , 144). So option B is ruled out.

Graph D : at t= 2 , the value of d is between d = 2890 and d = 3600. But our start point is (2 , 144). So option D is ruled out as well.

Please note that out interval is t E [2 , 5] and P and Q are the endpoints through which the secant line shall pass through.

=> P = (2 , 144) and Q (5 , 900)

And our distance function is d = 36t2 its a degree 2 function and the highest order coefficient is a positive that is 2.

So the fuction d(t) would have a shape of an upward parabola.

Now lets find the secant line from the using the point slope form :

We have two points (2,144) and (5,900)

so the slope of the secant line is , m = (900-144)/(5-2) = 252

Hence the line is : (y - y1) = m(t-t1)   , point slope form

y - 144 = 252(t-2) or y = 252t - 360

Hence the secant line passing through P and Q is : y = 252t - 360

In Grapg C : The secant line passes through the curve at (0,0) and (10,3600) and not the points P and Q

Hence Option C is ruled out as well.

Hence the correct graph is the graph in option A

b> From part (a) the slope of the secant line is = 252

This means the object falls at an average rate of 252 ft per seconds

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