Average Rate of Change Practice worksheet with problems using functions, graphs, and tables.

Let's solve each problem step by step.

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Problem 1:

Using the function $ f(x) = 2x^2 + 3x - 70 $ over the interval $[-2, 7]$

The average rate of change of a function over an interval $[a, b]$ is given by:

$$
\text{Average Rate of Change} = \frac{f(b) - f(a)}{b - a}
$$

Here, $ a = -2 $, $ b = 7 $

#### Step 1: Find $ f(-2) $
$$
f(-2) = 2(-2)^2 + 3(-2) - 70 = 2(4) - 6 - 70 = 8 - 6 - 70 = -68
$$

#### Step 2: Find $ f(7) $
$$
f(7) = 2(7)^2 + 3(7) - 70 = 2(49) + 21 - 70 = 98 + 21 - 70 = 49
$$

#### Step 3: Compute average rate of change
$$
\frac{f(7) - f(-2)}{7 - (-2)} = \frac{49 - (-68)}{7 + 2} = \frac{117}{9} = 13
$$

✔ Answer: $ \boxed{13.0} $

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Problem 2:

Find the average rate of change from $-3 \leq x \leq 9$ using the graph of $ f(x) $

We need to estimate values from the graph.

From the graph:
- At $ x = -3 $, $ f(-3) = 6 $
- At $ x = 9 $, $ f(9) = 6 $

Wait — let’s analyze carefully.

Looking at the graph:
- It appears to be a V-shaped graph, likely an absolute value function.
- The vertex is at $ (3, -4) $
- From $ x = -3 $ to $ x = 9 $, it goes from a high point down to the vertex and then up.

Estimate points:
- At $ x = -3 $: $ f(-3) = 6 $
- At $ x = 9 $: $ f(9) = 6 $

So both endpoints have $ y = 6 $

Then:
$$
\text{Average Rate of Change} = \frac{f(9) - f(-3)}{9 - (-3)} = \frac{6 - 6}{12} = \frac{0}{12} = 0
$$

✔ Answer: $ \boxed{0.0} $

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Problem 3:

Find the average rate of change over the interval $[-5, 15]$ using the table for $ g(x) $

We are given:

| x | g(x) |
|----|------|
| -3 | -5 |
| -5 | 20 |
| 10 | 15 |
| 15 | -2 |

We need values at $ x = -5 $ and $ x = 15 $

From the table:
- $ g(-5) = 20 $
- $ g(15) = -2 $

Now compute:
$$
\frac{g(15) - g(-5)}{15 - (-5)} = \frac{-2 - 20}{20} = \frac{-22}{20} = -1.1
$$

✔ Answer: $ \boxed{-1.1} $

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Problem 4:

The function $ b(t) = 20(1.5)^t + 5.2 $ represents tree frog population $ t $ years since 1990.

Find average rate of change between 1995 and 1997

First, convert years to $ t $:
- 1995 → $ t = 5 $
- 1997 → $ t = 7 $

Now compute $ b(5) $ and $ b(7) $

#### Step 1: $ b(5) $
$$
b(5) = 20(1.5)^5 + 5.2
$$
Calculate $ (1.5)^5 $:
$$
1.5^2 = 2.25 \\
1.5^4 = (2.25)^2 = 5.0625 \\
1.5^5 = 5.0625 \times 1.5 = 7.59375
$$
So:
$$
b(5) = 20(7.59375) + 5.2 = 151.875 + 5.2 = 157.075
$$

#### Step 2: $ b(7) $
$$
b(7) = 20(1.5)^7 + 5.2
$$
We already have $ 1.5^5 = 7.59375 $
Then:
- $ 1.5^6 = 7.59375 \times 1.5 = 11.390625 $
- $ 1.5^7 = 11.390625 \times 1.5 = 17.0859375 $

So:
$$
b(7) = 20(17.0859375) + 5.2 = 341.71875 + 5.2 = 346.91875
$$

#### Step 3: Average rate of change
$$
\frac{b(7) - b(5)}{7 - 5} = \frac{346.91875 - 157.075}{2} = \frac{189.84375}{2} = 94.921875
$$

Round to nearest hundredth:
$$
\boxed{94.92}
$$

✔ Answer: $ \boxed{94.92} $

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✔ Final Answers Summary:

1. $ 2x^2 + 3x - 70 $ on $[-2, 7]$: 13.0
2. Graph from $-3 \leq x \leq 9$: 0.0
3. Table $ g(x) $ on $[-5, 15]$: -1.1
4. Frog population 1995–1997: 94.92

Let me know if you'd like these explained in more detail!