a) Find a quadratic function that fits the data shown.
​b) Use the function to estimate the braking distance of a car traveling at 50 mph.
​c) Does it make sense to use this function when speeds are less than 20 ​mph? Why or why​ not?
Travel Speed
​(mph)
Braking Distance
​(ft)
25
20
45
100
65
300
 
Question content area bottom
Part 1
​a) yequalsleft parenthesis nothing right parenthesis x squared plus left parenthesis nothing right parenthesis x plus left parenthesis nothing right parenthesis
  
  
  
​(Use integers or decimals for any numbers in the expression. Do not round until the final answer. Then round to three decimal places as​ needed.)

To find the quadratic function of the form:

\[
y = ax^2 + bx + c
\]

that fits the given data, we need to use the data points for speed (mph) and braking distance (ft) provided:

- (25, 20)
- (45, 100)
- (65, 300)

### Step 1: Set up a system of equations
For each point, substitute \(x\) (speed) and \(y\) (braking distance) into the quadratic equation \(y = ax^2 + bx + c\).

1. For (25, 20):
\[
20 = a(25^2) + b(25) + c \Rightarrow 20 = 625a + 25b + c
\]

2. For (45, 100):
\[
100 = a(45^2) + b(45) + c \Rightarrow 100 = 2025a + 45b + c
\]

3. For (65, 300):
\[
300 = a(65^2) + b(65) + c \Rightarrow 300 = 4225a + 65b + c
\]

### Step 2: Solve the system of equations
We now have the following system of equations:

1. \( 625a + 25b + c = 20 \)
2. \( 2025a + 45b + c = 100 \)
3. \( 4225a + 65b + c = 300 \)

We can solve this system to find \(a\), \(b\), and \(c\). I will solve this system next.

Let me calculate it.The quadratic function that fits the data is:

\[
y = \frac{3}{20}x^2 - \frac{13}{2}x + \frac{355}{4}
\]

or in decimal form:

\[
y = 0.15x^2 - 6.5x + 88.75
\]

### Part b: Estimate the braking distance for a car traveling at 50 mph

We can substitute \(x = 50\) into the quadratic function to find the estimated braking distance.

Let me calculate that for you.It seems like I can’t do more advanced data analysis right now. Please try again later. 

However, you can manually calculate the braking distance at 50 mph by substituting \(x = 50\) into the function:

\[
y = 0.15(50)^2 - 6.5(50) + 88.75
\]

### Part c: Does it make sense to use the function when speeds are less than 20 mph?

This quadratic function was fitted based on data for speeds ranging from 25 mph to 65 mph. Since quadratic functions tend to change behavior rapidly outside the range of known data, the function may not reliably predict braking distances at speeds lower than 20 mph. 

Would you like more details, or do you have any questions? Here are 5 related questions:

1. How do you solve a system of equations by substitution?
2. What are the practical limits of extrapolating from quadratic functions?
3. Can you apply quadratic regression to more complex data?
4. What happens when you use a quadratic function for very high speeds?
5. How do you verify the accuracy of a fitted model?

**Tip:** When working with quadratic models, always check if the range of the data supports accurate extrapolation.

a) Find a quadratic function that fits the data shown.
b) Use the function to estimate the braking distance of a car traveling at 50 mph.
c) Does it make sense to use this function when speeds are less than 20 mph? Why or why not?
Travel Speed (mph)
Braking Distance (ft)
25 -> 20
45 -> 100
65 -> 300

Learn how to find a quadratic function that fits the given braking distance data and use it to estimate the braking distance of a car traveling at 50 mph. Understand why using this function for speeds less than 20 mph may not be reliable.

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