Question

the price of a ticket to a baseball game can be approximated by $ p(x) = 0.03x^2 + 0.58x + 8.75 $, where $ x $ is the number of years after 1991 and $ p(x) $ is in dollars
a) find the price of a ticket in the year 2021
b) at what rate will the ticket prices be changing in the year 2021?

a) $ \square $ (round your answer to the nearest cent)
b) $ \square $ (round your answer to the nearest cent )

Explanation:

Part (a)

Step1: Find the value of \( x \)

To find \( x \) for the year 2021, we calculate the number of years after 1991. So, \( x = 2021 - 1991 = 30 \).

Step2: Substitute \( x = 30 \) into \( p(x) \)

The function is \( p(x)=0.03x^{2}+0.58x + 8.75 \). Substituting \( x = 30 \):
\[

$$\begin{align*} p(30)&=0.03\times(30)^{2}+0.58\times30 + 8.75\\ &=0.03\times900+17.4 + 8.75\\ &=27+17.4+8.75\\ &=53.15 \end{align*}$$

\]

Step1: Find the derivative of \( p(x) \)

The function \( p(x)=0.03x^{2}+0.58x + 8.75 \). Using the power rule, the derivative \( p'(x) \) is:
\( p'(x)=2\times0.03x+0.58 = 0.06x + 0.58 \)

Step2: Substitute \( x = 30 \) into \( p'(x) \)

Substituting \( x = 30 \) into \( p'(x) \):
\[

$$\begin{align*} p'(30)&=0.06\times30+0.58\\ &=1.8 + 0.58\\ &=2.38 \end{align*}$$

\]

Answer:

\( \$53.15 \)

Part (b)