Question

The average price of a ticket to a baseball game can be approximated by ​p(x)equals0.03 x squared plus 0.42 x plus 9.63​, where x is the number of years after 1991 and​ p(x) is in dollars. ​a) Find ​p(3​). ​b) Find ​p(13​). ​c) Find ​p(13​)minus​p(3​). ​d) Find StartFraction p (13 )minus p (3 )Over 13 minus 3 EndFraction ​, and interpret this result.

Answer 1

Answer:

a) We want to find $p(3)$ and if we replace x=3 into p(x) we got:

$p(3) = 0.03*(3)^2 +0.42*3 +9.63 = 11.16$

b) We want to find $p(13)$ and if we replace x=13 into p(x) we got:

$p(13) = 0.03*(13)^2 +0.42*13 +9.63 = 20.16$

c) $p(13)-p(3) = [0.03*(13)^2 +0.42*13 +9.63]-[0.03*(3)^2 +0.42*3 +9.63] = 20.16-11.16 = 9$

d)  $\frac{p(13)-p(3)}{13-3}$

If we use the result from part c we have:

$\frac{p(13)-p(3)}{13-3}= \frac{9}{10}= 0.9$

And the interpretation for this case would be:

It represents the average rate of change in price from 1996 to 2006.

Step-by-step explanation:

For this case we have the following function given:

$p(x) = 0.03 x^2 +0.42 x +9.63$

Part a

We want to find $p(3)$ and if we replace x=3 into p(x) we got:

$p(3) = 0.03*(3)^2 +0.42*3 +9.63 = 11.16$

Part b

We want to find $p(13)$ and if we replace x=13 into p(x) we got:

$p(13) = 0.03*(13)^2 +0.42*13 +9.63 = 20.16$

Part c

For this case we want to find $p(13) -p(3)$ and we have from the results of part a and b this:

$p(13)-p(3) = [0.03*(13)^2 +0.42*13 +9.63]-[0.03*(3)^2 +0.42*3 +9.63] = 20.16-11.16 = 9$

Part d

For this case we want to find:

$\frac{p(13)-p(3)}{13-3}$

If we use the result from part c we have:

$\frac{p(13)-p(3)}{13-3}= \frac{9}{10}= 0.9$

And the interpretation for this case would be:

It represents the average rate of change in price from 1996 to 2006.