Calculus worksheet featuring graphs and problems on average rates of change and secant lines.

Let's solve each problem step by step, starting with the average rate of change, then moving on to the equation of the secant line.

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🔹 Problem 1:

Find the average rate of change of $ y = x^2 - x + 1 $ over $[0, 3]$

#### ✔ Step 1: Use the formula for average rate of change:
$$
\text{Average rate of change} = \frac{f(b) - f(a)}{b - a}
$$
Here, $ a = 0 $, $ b = 3 $

#### ✔ Step 2: Evaluate $ f(0) $ and $ f(3) $
- $ f(0) = (0)^2 - (0) + 1 = 1 $
- $ f(3) = (3)^2 - (3) + 1 = 9 - 3 + 1 = 7 $

#### ✔ Step 3: Plug into formula
$$
\frac{f(3) - f(0)}{3 - 0} = \frac{7 - 1}{3} = \frac{6}{3} = 2
$$

✔ Answer: 2

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🔹 Problem 2:

Find the average rate of change of $ y = -\frac{1}{x - 2} $ over $[-3, -2]$

#### ✔ Step 1: Use the same formula:
$$
\frac{f(-2) - f(-3)}{-2 - (-3)} = \frac{f(-2) - f(-3)}{1}
$$

#### ✔ Step 2: Evaluate $ f(-2) $ and $ f(-3) $
- $ f(-2) = -\frac{1}{-2 - 2} = -\frac{1}{-4} = \frac{1}{4} $
- $ f(-3) = -\frac{1}{-3 - 2} = -\frac{1}{-5} = \frac{1}{5} $

#### ✔ Step 3: Compute difference
$$
\frac{1/4 - 1/5}{1} = \frac{5 - 4}{20} = \frac{1}{20}
$$

✔ Answer: $ \frac{1}{20} $

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🔹 Problem 3:

Find the equation of the secant line that intersects the points $(-2, -4)$ and $(1, 2)$ on the function $ y = -x^2 + x + 2 $

> Note: The function is given, but we are told the two points lie on it. We can verify this, but the question asks only for the secant line, so we use the two points.

#### ✔ Step 1: Find the slope $ m $
$$
m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{2 - (-4)}{1 - (-2)} = \frac{6}{3} = 2
$$

#### ✔ Step 2: Use point-slope form
Use point $ (1, 2) $:
$$
y - 2 = 2(x - 1)
$$
Simplify:
$$
y = 2x - 2 + 2 = 2x
$$

✔ Answer: $ y = 2x $

> 🔍 Verification: Check if both points satisfy $ y = 2x $:
- $ x = -2 \Rightarrow y = -4 $ ✔
- $ x = 1 \Rightarrow y = 2 $ ✔

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🔹 Problem 4:

Find the equation of the secant line that intersects the points $ (1, -1) $ and $ \left(3, -\frac{1}{3}\right) $ on the function $ y = -\frac{1}{x} $

Again, use the two points to find the secant line.

#### ✔ Step 1: Find slope $ m $
$$
m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-\frac{1}{3} - (-1)}{3 - 1} = \frac{-\frac{1}{3} + 1}{2} = \frac{\frac{2}{3}}{2} = \frac{1}{3}
$$

#### ✔ Step 2: Use point-slope form
Use point $ (1, -1) $:
$$
y - (-1) = \frac{1}{3}(x - 1)
\Rightarrow y + 1 = \frac{1}{3}x - \frac{1}{3}
\Rightarrow y = \frac{1}{3}x - \frac{1}{3} - 1 = \frac{1}{3}x - \frac{4}{3}
$$

✔ Answer: $ y = \frac{1}{3}x - \frac{4}{3} $

> 🔍 Verification:
- At $ x = 1 $: $ y = \frac{1}{3}(1) - \frac{4}{3} = -1 $ ✔
- At $ x = 3 $: $ y = \frac{1}{3}(3) - \frac{4}{3} = 1 - \frac{4}{3} = -\frac{1}{3} $ ✔

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✔ Final Answers:

1) Average rate of change: 2
2) Average rate of change: $ \frac{1}{20} $
3) Secant line: $ y = 2x $
4) Secant line: $ y = \frac{1}{3}x - \frac{4}{3} $

Let me know if you'd like these graphed or explained further!