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.'

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!

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.'

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, the square root function only produces non-negative outputs—you can never get a negative number from a principal square root. The smallest output is √0 = 0, and as x increases, √x increases without bound. Choice B is correct because [0,∞) represents all non-negative real numbers, with the square bracket showing 0 is included as the minimum possible output. Choice C incorrectly suggests negative outputs are possible, but √x is always non-negative by definition. 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!

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!

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.'

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 C(n) = 5 + 12n where n is the number of tickets, n must be a non-negative integer because you can't buy a negative or fractional number of tickets in this context. Choice D is correct because it specifies all whole numbers starting from 0, which includes buying zero tickets (just the booking fee) up to any positive integer. Choice A might confuse you if you thought n could be any real number greater than or equal to zero, but tickets are discrete, so only integers make sense—always consider the real-world meaning! 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.'

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(x) = √(9 - x), the inside must be non-negative, so 9 - x ≥ 0, which solves to x ≤ 9. Choice B is correct because (-∞, 9] includes all x up to and including 9, where h(9) = √0 = 0 is defined. Choice C excludes 9 with a parenthesis, but that's not needed since x=9 works—double-check if the boundary value is allowed! 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.'

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-2), we need the expression under the square root to be non-negative, so x-2 ≥ 0, which means x ≥ 2. Choice B is correct because [2,∞) represents all real numbers greater than or equal to 2, using a square bracket at 2 to show it's included. Choice A would give negative values under the square root, making the output imaginary rather than real. 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.'

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 f(x) = x² - 4, this is a parabola opening upward with vertex at (0, -4), so the minimum y is -4, and it goes to infinity as x moves away from 0. Choice C is correct because [-4, ∞) captures all y ≥ -4, including -4 when x=0. Choice B is incorrect because it suggests y ≤ -4, but that's for a downward-opening parabola—remember to check if the coefficient of x² is positive (up) or negative (down)! 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!