Linear and Exponential Relationships
Unit 2 of Algebra I. 13 questions below, each with the working. Every answer was checked by a second pass before it was published.
Systems of linear equations and inequalities, functions and function notation, domain and range, sequences, linear versus exponential growth, graphing and transforming linear and exponential functions.
How this unit is tested
What you have to know
13 practice questions
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Solve the system: $y = 2x - 3$ and $y = -x + 9$.
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Answer. x = 4, y = 5
Since both equations equal y, set them equal: $2x-3=-x+9$. Adding x to both sides and adding 3 gives $3x=12$, so $x=4$. Substituting back gives $y=2(4)-3=5$. -
Solve the system by elimination: $3x + 2y = 16$ and $x - 2y = 0$.
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Answer. x = 4, y = 2
The y-terms are opposites, so adding the equations eliminates y: $4x = 16$, giving $x=4$. Substituting into $x-2y=0$ gives $4-2y=0$, so $y=2$. -
Which point satisfies both $y > x + 1$ and $y \le -2x + 6$?
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Answer. (1, 3)
Test each point in both inequalities. For (1,3): $3>1+1=2$ is true, and $3\le -2(1)+6=4$ is true, so it satisfies both. The other points each fail at least one inequality. -
If $f(x) = 2x^2 - 3x + 1$, find $f(-2)$.
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Answer. 15
Substitute -2 for every x: $2(-2)^2-3(-2)+1 = 2(4)+6+1 = 8+6+1=15$. -
A graph shows an upward-opening parabola with vertex at (2, -3), extending infinitely to the left and right. State its domain and range.
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Answer. Domain: all real numbers. Range: y ≥ -3.
Since the graph extends left and right without bound, every x-value is possible, giving domain of all real numbers. The vertex at y = -3 is the lowest point, and the graph opens upward, so the range is all y-values greater than or equal to -3. -
A vending machine's profit is modeled by $P(x) = 1.5x - 20$, where x is the number of snacks sold, and the machine holds at most 50 snacks. What is a reasonable domain for this function?
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Answer. x is a whole number with 0 ≤ x ≤ 50
Since x represents a count of snacks sold, it can't be negative or a fraction, and it can't exceed the machine's capacity of 50, so the domain is restricted to whole numbers from 0 to 50. -
An arithmetic sequence has first term $a_1 = 7$ and common difference $d = 4$. Find $a_{12}$.
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Answer. 51
Using $a_n = a_1 + (n-1)d$: $a_{12} = 7 + (12-1)(4) = 7 + 44 = 51$. -
A geometric sequence has first term $a_1 = 3$ and common ratio $r = 2$. Find $a_7$.
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Answer. 192
Using $a_n = a_1 \cdot r^{(n-1)}$: $a_7 = 3 \cdot 2^{6} = 3 \cdot 64 = 192$. -
A table shows x = 0, 1, 2, 3 with y = 4, 12, 36, 108. Is this pattern linear or exponential, and what is its equation?
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Answer. Exponential; $y = 4(3)^x$
The differences between y-values (8, 24, 72) are not constant, but the ratios (12/4=3, 36/12=3, 108/36=3) are constant, so the pattern is exponential with initial value 4 and growth factor 3. -
An exponential function passes through (0, 50) and (3, 6.25). Find its equation.
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Answer. $y = 50(0.5)^x$
The initial value is a = 50 (the y-value at x = 0). Using the second point: $50 r^3 = 6.25$, so $r^3 = 0.125$, giving $r = 0.5$. The equation is $y = 50(0.5)^x$, a decay function. -
Describe how the graph of $g(x) = \tfrac{1}{2}x - 4$ is transformed from the parent function $f(x) = x$.
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Answer. Vertically compressed by a factor of 1/2 (less steep), then shifted down 4 units
The slope changed from 1 to 1/2, which makes the line rise more slowly (a vertical compression), and the constant term of -4 moves the entire graph down 4 units compared to $f(x)=x$. -
Describe how the graph of $g(x) = -2^x + 5$ is transformed from the parent function $f(x) = 2^x$.
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Answer. Reflected over the x-axis, then shifted up 5 units; the horizontal asymptote moves from y = 0 to y = 5
The negative sign in front of $2^x$ flips the graph upside down (reflection over the x-axis), and adding 5 shifts the entire graph, including its horizontal asymptote, up from y = 0 to y = 5. -
A bacteria population starts at 200 and grows 15% each hour. Write a function for the population and find the population after 6 hours (round to the nearest whole bacterium).
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Answer. $P(t) = 200(1.15)^t$; about 463 bacteria after 6 hours
A 15% hourly increase means the growth factor is 1 + 0.15 = 1.15, and the initial value is 200, giving $P(t)=200(1.15)^t$. Evaluating at t = 6: $P(6) = 200(1.15)^6 \approx 200(2.313) \approx 463$.
What people get wrong
- Forgetting to flip the inequality symbol when multiplying or dividing both sides by a negative number while solving a system of inequalities — always check the sign of the coefficient before dividing.
- Treating f(-2) as -f(2), i.e. evaluating the function then negating it, instead of substituting -2 directly for every x in the rule — always replace the variable itself with the given input.
- Reading a graph's domain or range and including an endpoint that has an open circle — an open circle means that value is not included, so the inequality must be strict (< or >), not ≤ or ≥.
- Applying the geometric sequence formula (multiplying by r) to a sequence that is actually arithmetic, or vice versa — check whether consecutive terms have a constant difference or a constant ratio before choosing a formula.
- Assuming a table shows exponential growth just because the y-values are increasing quickly, without actually checking that consecutive ratios are constant — a fast-growing linear pattern can look similar at a glance.
- Mixing up the direction of a horizontal shift, writing $f(x-3)$ as a shift left instead of right — a subtraction inside the function argument shifts the graph right, not left.
Drill this unit until it sticks
These questions come back on a schedule built from what you get wrong, alongside the rest of Algebra I. Free, and no account needed to start.