Understanding Functions, Domain, and Range - Algebra 1

Which set represents the domain of the function $f(x)=\dfrac{x+1}{x^2-9}$ ? Give your answer in interval notation.
Answer: $( -\infty, -3)\cup(-3, 3)\cup(3, \infty)$ This question tests your understanding of what functions are, and how to determine their domains (possible inputs) and ranges (possible outputs). The domain is the set of all possible input values (x-values) that make sense for the function: for formulas, we exclude values that would cause division by zero or square roots of negatives, and in real-world contexts, we only include values that are realistic (like you can't have -3 people or 2.5 items if they're discrete). For f(x) = (x+1)/(x² - 9), factor the denominator as (x-3)(x+3), so it's undefined at x=3 and x=-3 where denominator is zero. Choice A is correct because (-∞, -3) ∪ (-3, 3) ∪ (3, ∞) excludes exactly those two points. Choice C excludes only 3, but you need to catch both roots of the denominator—always factor to find all zeros! For finding domain from a formula: (1) Start by assuming all real numbers are okay, (2) Then look for restrictions—is there a square root (need inside ≥ 0)? A fraction (need denominator ≠ 0)? (3) Write the domain excluding or including only the values that work. Most Algebra 1 functions have domains like 'all reals' or 'x ≥ some number' or 'all reals except one value.'
A function $f$ satisfies $f(2) = 7$ , $f(5) = 7$ , and $f(8) = 12$ . Based on this information alone, which statement must be true?
Answer: The value 7 is in the range of $f$ and the function is not one-to-one over the given points When you encounter a problem giving you specific function values, you're being tested on fundamental concepts about functions, domains, ranges, and the relationships between inputs and outputs. Let's analyze what the given information tells us. We know $f(2) = 7$ , $f(5) = 7$ , and $f(8) = 12$ . This means that when we input 2 or 5, we get the same output of 7, while inputting 8 gives us 12. This immediately tells us two important things: the value 7 is definitely in the range of $f$ (since it's an output), and the function is not one-to-one over these points because two different inputs (2 and 5) produce the same output (7). This makes choice D correct. Now let's examine why the other choices are wrong. Choice A incorrectly states that 7 is in the domain - but 7 is an output value, so it's in the range, not the domain. The domain consists of the input values: 2, 5, and 8. Choice B makes an unjustified assumption about the complete function. We only know information about three specific points, but the function could have many more elements in both its domain and range. Choice C reflects a common misconception - functions are absolutely allowed to have multiple inputs produce the same output. What's forbidden is one input producing multiple outputs. Remember: functions can be many-to-one (multiple inputs giving the same output) but never one-to-many. When analyzing function problems, carefully distinguish between domain (inputs) and range (outputs), and don't assume the given information represents the entire function unless explicitly stated.
For what values of $x$ is the function $g(x)=\dfrac{3}{x-2}$ defined? Write your answer in words or using interval notation.
Answer: All real numbers except $x=2$ This question tests your understanding of what functions are, and how to determine their domains (possible inputs) and ranges (possible outputs). The domain is the set of all possible input values (x-values) that make sense for the function: for formulas, we exclude values that would cause division by zero or square roots of negatives, and in real-world contexts, we only include values that are realistic (like you can't have -3 people or 2.5 items if they're discrete). For g(x) = 3/(x - 2), the function is undefined when the denominator is zero, so $x - 2 = 0$ means $x = 2$ is excluded, but all other real numbers work. Choice A is correct because it states all real numbers except $x=2$ , which matches the restriction from the denominator. Choice B might seem tempting if you mistakenly thought the numerator caused an issue, but the numerator is just 3, a constant, so no problem there—always check the denominator for zeros! For finding domain from a formula: (1) Start by assuming all real numbers are okay, (2) Then look for restrictions—is there a square root (need inside $\geq 0$ )? A fraction (need denominator $\neq 0$ )? (3) Write the domain excluding or including only the values that work. Most Algebra 1 functions have domains like 'all reals' or ' $x \geq$ some number' or 'all reals except one value.'
Let $q(x)=\dfrac{x+1}{x+2}$ . For each input $x$ in the domain, the function must produce exactly one output, so division by zero is not allowed. What is the domain of $q$ ? (Use interval notation.)
Answer: $(-\infty, -2) \cup (-2, \infty)$ This question tests your understanding of what functions are, and how to determine their domains (possible inputs) and ranges (possible outputs). The domain is the set of all possible input values (x-values) that make sense for the function: for formulas, we exclude values that would cause division by zero or square roots of negatives, and in real-world contexts, we only include values that are realistic (like you can't have -3 people or 2.5 items if they're discrete). For q(x) = (x+1)/(x+2), the denominator can't be zero, so x ≠ -2, excluding just that point from all reals. Choice A is correct because (-∞, -2) ∪ (-2, ∞) properly leaves out only x=-2. Choice C excludes x=2 instead, but the denominator is x+2, not x-2—gently double-check where the denominator is zero. For finding domain from a formula: (1) Start by assuming all real numbers are okay, (2) Then look for restrictions—is there a square root (need inside ≥ 0)? A fraction (need denominator ≠ 0)? (3) Write the domain excluding or including only the values that work. Most Algebra 1 functions have domains like 'all reals' or 'x ≥ some number' or 'all reals except one value.'
The function $k$ is defined by $k(x)=\sqrt{9-x}$ . What is the domain of $k$ ? (Give your answer in interval notation.)
Answer: $( -\infty,9]$ This question tests your understanding of what functions are, and how to determine their domains (possible inputs) and ranges (possible outputs). The domain is the set of all possible input values (x-values) that make sense for the function: for formulas, we exclude values that would cause division by zero or square roots of negatives, and in real-world contexts, we only include values that are realistic (like you can't have -3 people or 2.5 items if they're discrete). For k(x) = √(9-x), we need 9 - x ≥ 0, which means 9 ≥ x, or equivalently x ≤ 9. Choice A is correct because (-∞,9] represents all real numbers less than or equal to 9, with the square bracket showing 9 is included since √0 = 0 is valid. Choice B would be the domain if we had √(x-9) instead, requiring x ≥ 9. For finding domain from a formula: (1) Start by assuming all real numbers are okay, (2) Then look for restrictions—is there a square root (need inside ≥ 0)? A fraction (need denominator ≠ 0)? (3) Write the domain excluding or including only the values that work. When the variable is subtracted inside a square root, flip your thinking!
The height of a toy rocket above the ground is modeled by $H(t)=-2t^2+8t+1$ , where $t$ is time in seconds. The model is only valid from launch until it hits the ground at $t=4$ . What is the domain of $H$ in this context?
Answer: $[0,4]$ This question tests your understanding of what functions are, and how to determine their domains (possible inputs) and ranges (possible outputs). The domain is the set of all possible input values (x-values) that make sense for the function: for formulas, we exclude values that would cause division by zero or square roots of negatives, and in real-world contexts, we only include values that are realistic (like you can't have -3 people or 2.5 items if they're discrete). For the rocket height H(t) = -2t² + 8t + 1, the context limits t from launch at t=0 to when it hits the ground at t=4, so only those times are meaningful. Choice C is correct because [0,4] includes t=0 (launch) and t=4 (landing), capturing the valid interval. Choice A ignores the context and assumes all real t, but negative time or after landing doesn't make sense here—always factor in the real-world scenario! For finding domain from a formula: (1) Start by assuming all real numbers are okay, (2) Then look for restrictions—is there a square root (need inside ≥ 0)? A fraction (need denominator ≠ 0)? (3) Write the domain excluding or including only the values that work. Most Algebra 1 functions have domains like 'all reals' or 'x ≥ some number' or 'all reals except one value.'
The function $f$ is defined by $f(x)=\sqrt{x-4}$ . What is the domain of $f$ ? (Give your answer in interval notation.)
Answer: $[4,\infty)$ This question tests your understanding of what functions are, and how to determine their domains (possible inputs) and ranges (possible outputs). The domain is the set of all possible input values (x-values) that make sense for the function: for formulas, we exclude values that would cause division by zero or square roots of negatives, and in real-world contexts, we only include values that are realistic (like you can't have -3 people or 2.5 items if they're discrete). For f(x) = √(x-4), we need the expression inside the square root to be non-negative, so x - 4 ≥ 0, which means x ≥ 4. Choice B is correct because [4,∞) includes all real numbers greater than or equal to 4, with the square bracket showing that 4 itself is included since √0 = 0 is valid. Choice A would allow negative values under the square root, which isn't allowed in real numbers. For finding domain from a formula: (1) Start by assuming all real numbers are okay, (2) Then look for restrictions—is there a square root (need inside ≥ 0)? A fraction (need denominator ≠ 0)? (3) Write the domain excluding or including only the values that work. Most Algebra 1 functions have domains like 'all reals' or 'x ≥ some number' or 'all reals except one value.'
A relation assigns to each input exactly one output, but assigns the same output to multiple different inputs. Which statement about this relation is necessarily true?
Answer: The relation is a function, and its range contains fewer elements than its domain contains Since each input gets exactly one output, this satisfies the definition of a function. Since some outputs correspond to multiple inputs, there must be fewer distinct outputs (range elements) than inputs (domain elements). Choice B incorrectly assumes equal cardinality. Choice C incorrectly thinks functions cannot have repeated outputs. Choice D misunderstands that the given condition already ensures it's a function.
The function $g$ is defined by $g(x)=\dfrac{5}{x-2}$ . For what values of $x$ is $g(x)$ defined? (Give your answer in words or set description.)
Answer: All real numbers except $x=2$ This question tests your understanding of what functions are, and how to determine their domains (possible inputs) and ranges (possible outputs). The domain is the set of all possible input values (x-values) that make sense for the function: for formulas, we exclude values that would cause division by zero or square roots of negatives, and in real-world contexts, we only include values that are realistic (like you can't have -3 people or 2.5 items if they're discrete). For g(x) = 5/(x-2), we need to avoid division by zero, which happens when the denominator x - 2 = 0, so x = 2 must be excluded from the domain. Choice A is correct because it states that g(x) is defined for all real numbers except x = 2, which is exactly where the denominator becomes zero. Choice B incorrectly excludes x = 5, but plugging in x = 5 gives g(5) = 5/3, which is perfectly valid. For finding domain from a formula: (1) Start by assuming all real numbers are okay, (2) Then look for restrictions—is there a square root (need inside ≥ 0)? A fraction (need denominator ≠ 0)? (3) Write the domain excluding or including only the values that work. When you see a fraction, always check where the denominator equals zero—those are your excluded values!
What are the possible output values (the range) of the function $p(x)=\sqrt{x}$ ? Give your answer in interval notation.
Answer: $[0, \infty)$ This question tests your understanding of what functions are, and how to determine their domains (possible inputs) and ranges (possible outputs). The range is the set of all possible output values (y-values) the function can produce: for a quadratic that opens upward with vertex at (2, -3), the range is all y-values greater than or equal to -3 because the parabola goes up from that minimum point forever. For p(x) = √x, inputs are x ≥ 0, and outputs start at 0 (when x=0) and go to infinity as x increases, never negative. Choice C is correct because [0, ∞) includes 0 and all positive reals, matching the non-negative outputs of square roots. Choice B suggests negative outputs, but square roots can't produce negatives in real numbers—think about how √x is always ≥ 0! For finding range: linear functions usually have range = all real numbers. For quadratics, find the vertex first—if it opens up, range is y ≥ (vertex y-value); if it opens down, range is y ≤ (vertex y-value). For square roots, range is usually y ≥ 0. The function type tells you a lot!
Consider the function $h(x)=x^2-4$ . What is the range of $h$ ? (Answer in interval notation.)
Answer: $[-4,\infty)$ This question tests your understanding of what functions are, and how to determine their domains (possible inputs) and ranges (possible outputs). The range is the set of all possible output values (y-values) the function can produce: for a quadratic that opens upward with vertex at (2, -3), the range is all y-values greater than or equal to -3 because the parabola goes up from that minimum point forever. For h(x) = x² - 4, this is a parabola opening upward (positive coefficient on x²) with vertex at (0, -4), found by noting the minimum occurs when x=0, giving h(0)=0-4=-4. Choice C is correct because [-4,∞) represents all y-values greater than or equal to -4, using a square bracket to show -4 is included as the minimum value. Choice B incorrectly suggests the range goes down to negative infinity, but parabolas opening upward have a minimum point, not a maximum. For finding range: linear functions usually have range = all real numbers. For quadratics, find the vertex first—if it opens up, range is y ≥ (vertex y-value); if it opens down, range is y ≤ (vertex y-value). For square roots, range is usually y ≥ 0. The function type tells you a lot!
The table shows some input-output pairs for function $k$ . If $k(x) = ax + b$ for constants $a$ and $b$ , what is $k(0)$ ?
Answer: 5 From the table, k(1) = 3 and k(3) = -1. Using k(x) = ax + b: k(1) = a + b = 3 and k(3) = 3a + b = -1. Subtracting: 2a = -4, so a = -2. Then b = 3 - a = 3 - (-2) = 5. Therefore k(0) = -2(0) + 5 = 5. Choice A uses k(3). Choice B uses the slope. Choice D adds k(1) and k(3) incorrectly.
Let $g(x)=\dfrac{1}{x-5}$ . To keep $g$ a function with one output for each input, $g(x)$ must be defined for the input. For what values of $x$ is $g(x)$ defined? (Answer in words or interval notation.)
Answer: All real numbers except $x=5$ This question tests your understanding of what functions are, and how to determine their domains (possible inputs) and ranges (possible outputs). The domain is the set of all possible input values (x-values) that make sense for the function: for formulas, we exclude values that would cause division by zero or square roots of negatives, and in real-world contexts, we only include values that are realistic (like you can't have -3 people or 2.5 items if they're discrete). For g(x) = 1/(x-5), we need to avoid division by zero, so the denominator x-5 cannot equal 0, which means x cannot equal 5. Choice A is correct because it states 'all real numbers except x=5', which excludes only the problematic value while allowing all others. Choice B incorrectly excludes x=0, but plugging in x=0 gives g(0)=1/(-5)=-1/5, which is perfectly defined. For finding domain from a formula: (1) Start by assuming all real numbers are okay, (2) Then look for restrictions—is there a square root (need inside ≥ 0)? A fraction (need denominator ≠ 0)? (3) Write the domain excluding or including only the values that work. Most Algebra 1 functions have domains like 'all reals' or 'x ≥ some number' or 'all reals except one value.'
The function $p$ is defined by $p(x)=-(x-1)^2+4$ . What is the range of $p$ ? (Give your answer in interval notation.)
Answer: $(-\infty,4]$ This question tests your understanding of what functions are, and how to determine their domains (possible inputs) and ranges (possible outputs). The range is the set of all possible output values (y-values) the function can produce: for a quadratic that opens upward with vertex at $(2, -3)$ , the range is all y-values greater than or equal to $-3$ because the parabola goes up from that minimum point forever. For $p(x) = -(x-1)^2 + 4$ , the negative sign in front means this parabola opens downward, and the vertex form shows the vertex is at $(1, 4)$ , giving a maximum value of 4. Choice B is correct because $(-\infty,4]$ represents all y-values less than or equal to 4, with the square bracket showing 4 is included since $p(1) = 4$ . Choice A would suggest the range starts at 4 and goes up, but this parabola opens down from its maximum. For finding range: linear functions usually have range = all real numbers. For quadratics, find the vertex first—if it opens up, range is $y \geq$ (vertex y-value); if it opens down, range is $y \leq$ (vertex y-value). The negative coefficient flips everything!
A ball's height (in meters) is modeled by a function $H(t)$ , where $t$ is time in seconds. The model is only valid from when the ball is thrown until it lands: $0\le t\le 4$ . What is the domain of $H$ ? (Answer in interval notation.)
Answer: $[0,4]$ This question tests your understanding of what functions are, and how to determine their domains (possible inputs) and ranges (possible outputs). The domain is the set of all possible input values (x-values) that make sense for the function: for formulas, we exclude values that would cause division by zero or square roots of negatives, and in real-world contexts, we only include values that are realistic (like you can't have -3 people or 2.5 items if they're discrete). For H(t) modeling a ball's height where the model is valid from 0 ≤ t ≤ 4 (thrown at t=0, lands at t=4), the domain includes all time values from 0 to 4 seconds. Choice B is correct because [0,4] uses square brackets at both ends to show that both t=0 (when thrown) and t=4 (when it lands) are included in the domain. Choice A incorrectly uses parentheses (0,4), which would exclude the moments of throwing and landing, missing important parts of the ball's journey. For finding domain from a formula: (1) Start by assuming all real numbers are okay, (2) Then look for restrictions—is there a square root (need inside ≥ 0)? A fraction (need denominator ≠ 0)? (3) Write the domain excluding or including only the values that work. Most Algebra 1 functions have domains like 'all reals' or 'x ≥ some number' or 'all reals except one value.'
A ball's height (in meters) after $t$ seconds is modeled by a function $H(t)$ , and the model is only intended to be used for $0\le t\le 6$ . What is the domain of $H$ in this context? (Use interval notation.)
Answer: $[0,6]$ This question tests your understanding of what functions are, and how to determine their domains (possible inputs) and ranges (possible outputs). The domain is the set of all possible input values (x-values) that make sense for the function: for formulas, we exclude values that would cause division by zero or square roots of negatives, and in real-world contexts, we only include values that are realistic (like you can't have -3 people or 2.5 items if they're discrete). The problem states the model is only intended for 0 ≤ t ≤ 6, meaning we're tracking the ball's height from t = 0 seconds (when thrown) to t = 6 seconds (perhaps when it hits the ground). Choice B is correct because [0,6] uses square brackets to show both endpoints are included—the model works at t = 0 and at t = 6. Choice A uses parentheses, incorrectly excluding the endpoints, but the problem clearly states t can equal 0 and 6. When a problem gives you explicit restrictions like "for 0 ≤ t ≤ 6," that's your domain! Square brackets [ ] mean "include this endpoint," while parentheses ( ) mean "get close but don't include."
For what values of $x$ is the function $g(x)=\dfrac{3}{x-2}$ defined? Give your answer in interval notation.
Answer: $(-\infty,2)\cup(2,\infty)$ This question tests your understanding of what functions are, and how to determine their domains (possible inputs) and ranges (possible outputs). The domain is the set of all possible input values (x-values) that make sense for the function: for formulas, we exclude values that would cause division by zero or square roots of negatives, and in real-world contexts, we only include values that are realistic (like you can't have -3 people or 2.5 items if they're discrete). For g(x) = 3/(x-2), the function is undefined when the denominator is zero, so x - 2 = 0 means x=2 is excluded, but all other real numbers work. Choice A is correct because (-∞, 2) ∪ (2, ∞) includes everything except x=2, where the function would involve division by zero. If you chose B thinking it's all reals, that's understandable, but remember to always check for denominator zeros—it's a key restriction! For finding domain from a formula: (1) Start by assuming all real numbers are okay, (2) Then look for restrictions—is there a square root (need inside ≥ 0)? A fraction (need denominator ≠ 0)? (3) Write the domain excluding or including only the values that work. Most Algebra 1 functions have domains like 'all reals' or 'x ≥ some number' or 'all reals except one value.'
If $g(x) = 3x - 2$ and $g(a) = 10$ , what is the value of $g(a + 1)$ ?
Answer: 13 First find a: if g(a) = 10, then 3a - 2 = 10, so 3a = 12, thus a = 4. Then g(a + 1) = g(5) = 3(5) - 2 = 13. Choice A incorrectly adds 1 to g(a). Choice C uses g(3a) instead of g(a+1). Choice D uses g(a) + 3a incorrectly.
Which set represents the domain of $f(x)=\dfrac{x+1}{x^2-9}$ ? Give your answer in interval notation.
Answer: $(-\infty,-3)\cup(-3,3)\cup(3,\infty)$ This question tests your understanding of what functions are, and how to determine their domains (possible inputs) and ranges (possible outputs). The domain is the set of all possible input values (x-values) that make sense for the function: for formulas, we exclude values that would cause division by zero or square roots of negatives, and in real-world contexts, we only include values that are realistic (like you can't have -3 people or 2.5 items if they're discrete). For f(x) = (x+1)/(x²-9), factor the denominator as (x-3)(x+3), so it's undefined at x=3 and x=-3 where denom=0. Choice B is correct because (-∞,-3) ∪ (-3,3) ∪ (3,∞) excludes exactly those two points, allowing all other real x. If you selected C, that misses excluding -3—remember to find all roots of the denominator for full restrictions. For finding domain from a formula: (1) Start by assuming all real numbers are okay, (2) Then look for restrictions—is there a square root (need inside ≥ 0)? A fraction (need denominator ≠ 0)? (3) Write the domain excluding or including only the values that work. Most Algebra 1 functions have domains like 'all reals' or 'x ≥ some number' or 'all reals except one value.'
Let $r(x)=\dfrac{x+2}{x-1}$ . What is the domain of $r$ ? (Answer in interval notation.)
Answer: $(-\infty,1)\cup(1,\infty)$ This question tests your understanding of what functions are, and how to determine their domains (possible inputs) and ranges (possible outputs). The domain is the set of all possible input values (x-values) that make sense for the function: for formulas, we exclude values that would cause division by zero or square roots of negatives, and in real-world contexts, we only include values that are realistic (like you can't have -3 people or 2.5 items if they're discrete). For r(x) = (x+2)/(x-1), we need to avoid division by zero, so the denominator x-1 cannot equal 0, which means x cannot equal 1. The domain is all real numbers except x = 1. Choice A is correct because (-∞,1)∪(1,∞) represents all real numbers less than 1 (using a parenthesis to exclude 1) union with all real numbers greater than 1 (again excluding 1 with a parenthesis). Choice D incorrectly excludes x = -1, but r(-1) = (-1+2)/(-1-1) = 1/(-2) = -1/2, which is perfectly defined. For finding domain from a formula: (1) Start by assuming all real numbers are okay, (2) Then look for restrictions—is there a square root (need inside ≥ 0)? A fraction (need denominator ≠ 0)? (3) Write the domain excluding or including only the values that work. Most Algebra 1 functions have domains like 'all reals' or 'x ≥ some number' or 'all reals except one value.'

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