Answer:
$25,900 initially
$8,128 after 11 years
Step-by-step explanation:
One thing to note, is I'm assuming you meant: [tex]v(t)=25,900(0.90)^t[/tex] and not [tex]v(t)=25,900(0.90)^2[/tex], otherwise that would just be a constant function and not an exponential function.
There are a couple key features of exponential functions we can derive from simply analyzing the equation.
For now I will be using the form: [tex]y=a(b)^x\text{ where b} > 0, \text{b}\ne 1[/tex]
The y-intercept, is just the "a" value, which is because the y-intercept is simply when x=0, and if we raise a number to the power of zero, we get one (excluding zero as a base), so we get the form: [tex]y=a[/tex], so "a" is our y-intercept.
there's actually two type of exponential functions: exponential growth and exponential decay.
Exponential Decay:
Exponential decay, as the name implies, is when a function is decaying or decreasing. This occurs when [tex]0 < b < 1[/tex].
The decay rate can be represented as "r" in decimal form, where: [tex]b=1-r[/tex].
The reason for this is because "1" can be thought of as 100% and "r" is r%, so you're taking away r% or decaying the value by r%, also known as a decay rate.
Exponential Growth:
Exponential growth as the name implies, is when a function is growing or increasing. This occurs when b > 1
The growth rate can be represented as "r" in decimal form where: [tex]b=1+r[/tex]
The reason for this is similar to decay rate except instead of taking away r% or decaying, it's "giving r%" or growing.
The question doesn't actually require finding exponential growth or decay, but it's still an important key feature.
In this case, we really just need to find the y-intercept and plug in t=11.
The y-intercept in many graphs, not just exponential graphs, represents some sort of initial value, which in this case, is the initial value of the car. As stated above, this y-intercept is just the value in front, which is 25,900.
As for the value after 11 years, we simply need to plug in t=11 to get the following: [tex]v(11)=25,900(0.90)^{11}[/tex], from here we can use a calculator to simplify this: [tex]v(11)\approx25,900(0.31381059609)\approx 8,127.694438731[/tex], which rounds to 8,128
