https://file.aiquickdraw.com/mathbot/file/1726591167836kxublvaw.jpg

The given function is:

\[
f(x) = 0.0086x^2 + 0.671x - 1.97
\]

where \( f(x) \) represents the stopping distance (in feet) and \( x \) is the speed of the car (in mph). We are asked to find the fastest speed \( x \) such that the stopping distance is 27.5 feet.

### Step 1: Set up the equation

Set \( f(x) = 27.5 \):

\[
0.0086x^2 + 0.671x - 1.97 = 27.5
\]

### Step 2: Rearrange the equation

Move 27.5 to the left side:

\[
0.0086x^2 + 0.671x - 1.97 - 27.5 = 0
\]

This simplifies to:

\[
0.0086x^2 + 0.671x - 29.47 = 0
\]

### Step 3: Solve using the quadratic formula

The quadratic formula is:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

For the equation \( 0.0086x^2 + 0.671x - 29.47 = 0 \), the coefficients are:

- \( a = 0.0086 \)
- \( b = 0.671 \)
- \( c = -29.47 \)

Now, let's solve this.The solutions to the quadratic equation are approximately:

\[
x_1 = 31.34 \, \text{mph}
\]
\[
x_2 = -109.36 \, \text{mph}
\]

Since the speed \( x \) must be positive, the fastest speed at which the car can be traveling and still stop within 27.5 feet is approximately **31.34 mph**.

Would you like further clarification or details? Here are some related questions you might explore:

1. How does increasing the speed affect the stopping distance?
2. Can this quadratic equation have multiple realistic solutions in different contexts?
3. How does the discriminant affect the number and type of solutions?
4. What is the physical meaning of the negative root in this context?
5. How would the stopping distance change if the coefficients in the equation were different?

**Tip:** When solving quadratic equations, always check the physical feasibility of the solutions based on the context of the problem.

The function f(x) = 0.0086x^2 + 0.671x - 1.97 represents the stopping distance in feet while driving under normal conditions at a speed of x mph. Find the fastest speed a car could be traveling (rounded to the nearest mile per hour) and still be able to stop within 27.5 feet while driving under normal conditions.

Learn how to solve the quadratic equation f(x) = 0.0086x^2 + 0.671x - 1.97 to find the fastest speed a car can travel while stopping within 27.5 feet. This problem demonstrates real-world applications of quadratic equations, suitable for Grades 10-12.

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