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In parts​ (A)-(C), find the indicated quantity for y equals f left parenthesis x right parenthesis equals 6 minus x squared. ​(A)  StartFraction f left parenthesis 2 right parenthesis minus f left parenthesis 1 right parenthesis Over left parenthesis 2 right parenthesis minus left parenthesis 1 right parenthesis EndFraction
equals
  
enter your response here 

Let's solve the given problem step by step.

We are asked to find the expression for:

\[
\frac{f(2) - f(1)}{2 - 1}
\]

where the function \( f(x) = 6 - x^2 \).

### Step 1: Calculate \( f(2) \)

Substitute \( x = 2 \) into the function:

\[
f(2) = 6 - (2^2) = 6 - 4 = 2
\]

### Step 2: Calculate \( f(1) \)

Substitute \( x = 1 \) into the function:

\[
f(1) = 6 - (1^2) = 6 - 1 = 5
\]

### Step 3: Compute the difference

Now, substitute the values into the expression:

\[
\frac{f(2) - f(1)}{2 - 1} = \frac{2 - 5}{1} = \frac{-3}{1} = -3
\]

### Final Answer:

\[
\frac{f(2) - f(1)}{2 - 1} = -3
\]

Would you like more details on this or have further questions? Here are some related questions to expand your understanding:

1. What is the average rate of change of \( f(x) \) over the interval [1, 3]?
2. How do we interpret the value of \(\frac{f(2) - f(1)}{2 - 1}\) geometrically?
3. What are the intercepts of the graph of \( f(x) = 6 - x^2 \)?
4. Can you find the vertex of the parabola given by \( f(x) = 6 - x^2 \)?
5. How does the value of \( f'(x) \) (the derivative) relate to the slope calculation we just performed?

### Tip:

In problems like this, the expression \(\frac{f(a) - f(b)}{a - b}\) represents the **average rate of change** of the function over the interval \([a, b]\), which geometrically is the slope of the secant line connecting the points on the curve at \( x = a \) and \( x = b \).

In parts (A)-(C), find the indicated quantity for y equals f(x) = 6 - x^2. (A) f(2) - f(1) over (2) - (1) equals ?

Learn how to find the average rate of change for the function f(x) = 6 - x^2 over a given interval. This problem involves calculating f(2) - f(1) over the interval from x = 1 to x = 2. Detailed step-by-step solution provided for students in Grades 9-11.

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