Algebra I Function Notation Worksheet featuring exercises on evaluating functions and converting statements to coordinate points.

Let's solve the Algebra I Function Notation Worksheet step by step.

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Given Functions:

- $ g(x) = -3x + 1 $
- $ f(x) = x^2 + 7 $
- $ h(x) = \frac{12}{x} $
- $ j(x) = 2x + 9 $

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## Problem 1: Evaluate the following expressions

a. $ g(10) = ? $

Plug in $ x = 10 $ into $ g(x) $:

$$
g(10) = -3(10) + 1 = -30 + 1 = \boxed{-29}
$$

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b. $ f(3) = ? $

$$
f(3) = (3)^2 + 7 = 9 + 7 = \boxed{16}
$$

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c. $ h(-2) = ? $

$$
h(-2) = \frac{12}{-2} = \boxed{-6}
$$

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d. $ j(7) = ? $

$$
j(7) = 2(7) + 9 = 14 + 9 = \boxed{23}
$$

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e. $ h(a) = ? $

This is just substituting $ a $ for $ x $:

$$
h(a) = \frac{12}{a}
$$

So, $ \boxed{\frac{12}{a}} $

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f. Find $ x $ if $ g(x) = 16 $

We know $ g(x) = -3x + 1 $. Set it equal to 16:

$$
-3x + 1 = 16
$$
$$
-3x = 15
$$
$$
x = \boxed{-5}
$$

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g. Find $ x $ if $ h(x) = -2 $

$ h(x) = \frac{12}{x} = -2 $

$$
\frac{12}{x} = -2
$$
Multiply both sides by $ x $:
$$
12 = -2x
$$
$$
x = \boxed{-6}
$$

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h. Find $ x $ if $ f(x) = 23 $

$ f(x) = x^2 + 7 = 23 $

$$
x^2 = 23 - 7 = 16
$$
$$
x = \pm \sqrt{16} = \boxed{\pm 4}
$$

So, $ x = 4 $ or $ x = -4 $

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i. CHALLENGE: $ g(b + c) $

We substitute $ b + c $ into $ g(x) $:

$$
g(b + c) = -3(b + c) + 1 = -3b - 3c + 1
$$

Answer: $ \boxed{-3b - 3c + 1} $

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j. CHALLENGE: $ f(h(x)) $

This is a composition of functions: plug $ h(x) $ into $ f(x) $

First, $ h(x) = \frac{12}{x} $

Now compute $ f(h(x)) = f\left(\frac{12}{x}\right) $

$$
f\left(\frac{12}{x}\right) = \left(\frac{12}{x}\right)^2 + 7 = \frac{144}{x^2} + 7
$$

Answer: $ \boxed{\frac{144}{x^2} + 7} $

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## Problem 2: Translate statements into coordinate points

In function notation, $ f(a) = b $ means the point $ (a, b) $

a. $ f(-1) = 1 $

This means when $ x = -1 $, $ y = 1 $

Point: $ \boxed{(-1, 1)} $

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b. $ h(2) = 7 $

When $ x = 2 $, $ y = 7 $

Point: $ \boxed{(2, 7)} $

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c. $ g(1) = -1 $

When $ x = 1 $, $ y = -1 $

Point: $ \boxed{(1, -1)} $

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d. $ k(3) = 9 $

Even though $ k(x) $ wasn't defined, we can still interpret this as:

When $ x = 3 $, $ y = 9 $

Point: $ \boxed{(3, 9)} $

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✔ Final Answers Summary:

#### Problem 1:
a. $ g(10) = \boxed{-29} $
b. $ f(3) = \boxed{16} $
c. $ h(-2) = \boxed{-6} $
d. $ j(7) = \boxed{23} $
e. $ h(a) = \boxed{\frac{12}{a}} $
f. $ x = \boxed{-5} $
g. $ x = \boxed{-6} $
h. $ x = \boxed{\pm 4} $
i. $ g(b+c) = \boxed{-3b - 3c + 1} $
j. $ f(h(x)) = \boxed{\frac{144}{x^2} + 7} $

#### Problem 2:
a. $ \boxed{(-1, 1)} $
b. $ \boxed{(2, 7)} $
c. $ \boxed{(1, -1)} $
d. $ \boxed{(3, 9)} $

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