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MATH 150 Algebraic Equations Problem Set With Solutions (Fall 2025)|ALG 244 Algebra Summer2023 Solutions 7933

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ALG 244 Algebra Summer2023 Solutions 7933 MATH 150 Algebraic Equations Problem Set With Solutions (Fall 2025)










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MATH 150 | September 13, 2025



Aiden Lopez
Daniel Lee
MATH 150

Algebraic Equations Problem Set
With Solutions

Problem 1: A bacteria culture starts with 271 bacteria and grows exponentially at a rate of
3.2% per hour. Write the exponential function and find the population after 8 hours.

Solution:
Exponential growth model: P(t) = P₀ × (1 + r)^t
P(t) = 271 × (1 + 0.032)^t
P(t) = 271 × 1.032^t
After 8 hours:
P(8) = 271 × 1.032^8
P(8) = 271 × 1.287
P(8) = 349 bacteria
(Similar to homework problem #12)
Note: I remember Professor Daniel Lee mentioned this approach



Problem 2: Graph the quadratic function f(x) = -1(x + 1)² - 3 Find: a) Vertex b) Axis of
symmetry c) y-intercept d) Domain and range e) Direction of opening

Solution:
Given: f(x) = -1(x + 1)² - 3
This is in vertex form: f(x) = a(x - h)² + k
a) Vertex: (h, k) = (-1, -3)
b) Axis of symmetry: x = -1
c) y-intercept: Find f(0)
f(0) = -1(0 + 1)² - 3
f(0) = -1(1)² - 3
f(0) = -1(1) - 3
f(0) = -1 - 3 = -4
y-intercept: (0, -4)
d) Domain: All real numbers, (-∞, ∞)
Range: (-∞, -3]


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, MATH 150 | September 13, 2025



e) Direction: Opens downward (since a = -1 < 0)
ASCII Graph Sketch:
|
|
|
|
|
----------|----------
|
|
*
*|*
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Vertex at (-1, -3)



Problem 3: A cell phone plan costs 43.96 per month plus 2.35 per text message. If the total
cost is $396.09, find the number of text messages used.

Solution:
Let x = number of text messages
Setting up the equation:
Total Cost = Fixed Cost + Variable Cost
396.09 = 43.96 + 2.35x
Solving for x:
396.09 - 43.96 = 2.35x
352.13 = 2.35x
x = 352.13 ÷ 2.35
x = 149.8 text messages
~~Wrong attempt crossed out~~
Actually, let me try again:
Check: 43.96 + 2.35(149.8) = 43.96 + 352.03 = 395.99 ✓



Problem 4: Given f(x) = 4x + 9 and g(x) = x² - 3, find: a) (f ∘ g)(x) b) (g ∘ f)(x) c) (f ∘ g)(3)

Solution:


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