Instructions: Graph the following function. Then, state whether the function represents exponential growth or decay.y=11x𝑦=11𝑥This function is exponential

The figure shows the graph of an exponential decay function 𝑓. The coordinates of two of the points are labeled. If 𝑦=𝑓𝑥, what is the 𝑦-coordinate of the point on the graph where 𝑥=0 ?Responses404030302020

This is the graph of the exponential function f(x).-6-5-4-3-2-1123456-6-5-4-3-2-1123456xyThe function g(x) is defined by g(x)=2x–2.Decide whether each statement is true or false.TrueFalseOn the interval from x=0 to x=3, the average rate of change of f(x) is greater than the average rate of change of g(x). The y-intercept of f(x) is greater than the y-intercept of g(x).The value f(1) is less than the y-intercept of g(x).The functions f(x) and g(x) have the same x-intercept.The value f(2) is equal to the value g(2).Submit

Select the correct answer.Which graph shows an exponential function that nears a constant value as x approaches positive infinity and has a y-inte

Instructions: Determine whether graphs of each of the exponential functions would show vertical stretch or vertical compression and whether there is reflection over the x𝑥-axis.y=−2(3)x

Let’s do an example together that is an exponential decay model.ProblemThe cost of a new car is $32,000$32,000. It depreciates at a rate of 15%15% per year. This means that it loses 15%15% of its value each year.Draw a graph of the car’s value against time in years.Find the formula that gives the value of the car in terms of time.Find the value of the car when it is four years old. SolutionThis is an exponential decay function. Start by making a table of values. To fill in the values we start with when t=0𝑡=0. Then we multiply the value of the car by 85%85% for each passing year. (Since the car loses 15%15% of its value, it keeps 85%85% of its value. 100−15=85100−15=85.) Remember 85%=0.8585%=0.85.Time Value (Thousands)0032321127.227.22223.123.13319.719.74416.716.75514.214.2The general formula is y=a(b)x𝑦=𝑎(𝑏)𝑥. In this case y𝑦 is the value of the car, x𝑥 is the time in years, a=32,000𝑎=32,000 is the starting amount in thousands, and b=0.85𝑏=0.85 since we multiply the value in any year by this factor to get the value of the car in the following year. The formula for this problem is:y=𝑦= (( )x)𝑥Finally, to find the value of the car when it is four years old, we use x=4𝑥=4 in the formula. Remember the value is in thousands.y=32000(0.85)4=𝑦=32000(0.85)4= At 44 years old, we expect the car to be worth $$ CheckQuestion 2