On a coordinate plane, graph the linear function . Find both intercepts of the function (as points).
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Algebra Quiz
Practice Graph Linear And Quadratic Functions in Algebra with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.
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On a coordinate plane, graph the linear function y=−x+2. Find both intercepts of the function (as points).
This quiz focuses on Graph Linear And Quadratic Functions, giving you a quick way to practice the rules, question types, and explanations that matter most for Algebra.
Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.
On a coordinate plane, graph the linear function y=−x+2. Find both intercepts of the function (as points).
Explanation: This question tests your understanding of how to identify key features of linear functions and their graphs, specifically intercepts. In a linear function y = mx + b, the slope m tells you how steep the line is and whether it rises (m > 0) or falls (m < 0), while the y-intercept b tells you where the line crosses the y-axis at the point (0, b). To find the y-intercept of y = -x + 2, we substitute x = 0 into the equation: y = - (0) + 2 = 2, so the graph crosses the y-axis at (0, 2); for the x-intercept, we set y = 0 and solve 0 = -x + 2, which gives x = 2, so the point is (2, 0). Choice A is correct because it properly identifies the intercepts as (0, 2) and (2, 0) using substitution and solving, with accurate calculations. Great job identifying this! Choice B mixes up the coordinates: intercepts are points, so we write them as (x, y) pairs—this choice has the y-intercept as (2, 0) instead of (0, 2), but the x-value always comes first! Quick trick for intercepts: y-intercept is always easy—just let x = 0 and calculate! For x-intercepts, set y = 0 and solve. And remember: intercepts are points with two coordinates, so write them as (x, y), not just single numbers. For linear functions in the form y = mx + b, you can read two key features immediately: m is the slope (how steep the line is), and b is the y-intercept (where it crosses the y-axis at (0, b)). No calculation needed—they're right there in the equation!
For the linear function y=21x−4, what is the slope?
Explanation: This question tests your understanding of how to identify key features of linear functions and their graphs, specifically the slope. In a linear function y = mx + b, the slope m tells you how steep the line is and whether it rises (m > 0) or falls (m < 0), while the y-intercept b tells you where the line crosses the y-axis at the point (0, b). In the equation y = (1/2)x - 4, the slope is 1/2. This means for every 1 unit we move to the right, the graph goes up by 1/2 unit. A positive slope means the line rises from left to right, while a negative slope means it falls. Choice C is correct because it properly identifies the slope as 1/2 by reading it from the coefficient of x in the equation. Great job identifying this! Choice D confuses slope with y-intercept—an easy mix-up when you're learning! In y = mx + b, the number in front of x is the slope (m = 1/2), and the number by itself is the y-intercept (b = -4). For linear functions in the form y = mx + b, you can read two key features immediately: m is the slope (how steep the line is), and b is the y-intercept (where it crosses the y-axis at (0, b)). No calculation needed—they're right there in the equation!
Graph the quadratic function y=(x+1)2+2 on a coordinate plane. What is the vertex?
Explanation: This question tests your understanding of how to identify key features of quadratic functions and their graphs, specifically the vertex. For a quadratic function in the form y = a(x - h)² + k, the vertex is the point (h, k), which is the highest point if the parabola opens down (a < 0) or the lowest point if it opens up (a > 0). This quadratic is in vertex form y = 1(x + 1)² + 2, which is y = 1(x - (-1))² + 2, so we can read the vertex directly: it's (h, k) = (-1, 2); the vertex is the turning point—the very bottom of the parabola since a > 0! Choice B is correct because it properly identifies the vertex as (-1, 2) by reading it from the vertex form, with accurate values. Choice A makes an error in the vertex calculation, using the wrong sign for h: since it's (x + 1)², that's (x - (-1))², so h = -1, not positive 1. With quadratics in vertex form y = a(x - h)² + k, the vertex is right in front of you: it's (h, k). Watch out though—there's a minus sign in the form, so if you see (x - 3)², the h-value is positive 3, but if you see (x + 3)², the h-value is -3!
For the quadratic function y=2x2−8x+6, what is the y-intercept (as a point) to label on the graph?
Explanation: This question tests your understanding of how to identify key features of quadratic functions and their graphs, specifically the y-intercept. The y-intercept is where the graph crosses the y-axis, which always happens when x = 0: to find it, substitute x = 0 into your function and you'll get the point (0, y-value). To find the y-intercept of y = 2x² - 8x + 6, we substitute x = 0 into the equation: y = 2(0)² - 8(0) + 6 = 6; this means the graph crosses the y-axis at the point (0, 6); it's the easiest intercept to find—just plug in 0 for x! Choice B is correct because it properly identifies the y-intercept as (0, 6) using substitution, with accurate calculation. Choice A gives just the number 2 instead of the point (0, 6); remember: intercepts are points on the graph, not just single numbers, so we need both coordinates! Quick trick for intercepts: y-intercept is always easy—just let x = 0 and calculate! And remember: intercepts are points with two coordinates, so write them as (x, y), not just single numbers.
On a coordinate plane, graph the quadratic function y=x2−5x+6. What are the x-intercepts (as points)?
Explanation: This question tests your understanding of how to identify key features of quadratic functions and their graphs, specifically x-intercepts. The x-intercept is where the graph crosses the x-axis, which always happens when y = 0: to find it, set your function equal to 0 and solve for x, giving you the point(s) (x-value, 0). To find the x-intercepts, we set y = 0 and solve: 0 = x² - 5x + 6. Factoring gives us (x - 2)(x - 3) = 0, so x = 2 and x = 3, giving us the points (2, 0) and (3, 0). The x-intercepts show where the graph touches or crosses the x-axis. Choice B is correct because it properly identifies the x-intercepts as (2, 0) and (3, 0) by factoring and solving correctly. Great job identifying this! Choice C makes an arithmetic mistake when factoring, using -2 and -3 instead of +2 and +3—remember to check which signs give the middle term -5x and product +6! Quick trick for intercepts: y-intercept is always easy—just let x = 0 and calculate! For x-intercepts, set y = 0 and solve. And remember: intercepts are points with two coordinates, so write them as (x, y), not just single numbers.
On a coordinate plane, the quadratic function y=x2−5x+6 is graphed. What are the x-intercepts of y=x2−5x+6 (as points)?
Explanation: This question tests your understanding of how to identify key features of quadratic functions and their graphs, specifically x-intercepts. The x-intercept is where the graph crosses the x-axis, which always happens when y = 0: to find it, set your function equal to 0 and solve for x, giving you the point(s) ([x-value], 0). To find the x-intercept(s), we set y = 0 and solve: 0 = x² - 5x + 6; factoring gives us (x - 2)(x - 3) = 0, so x = 2 and x = 3, giving us the points (2, 0) and (3, 0)—the x-intercepts show where the graph touches or crosses the x-axis. Choice A is correct because it properly identifies the x-intercepts as (2, 0) and (3, 0) by factoring and solving accurately. Great job identifying this! Choice C makes an arithmetic mistake when factoring, using -2 and -3 instead of 2 and 3; we all make sign errors—that's why checking our work is so important! Quick trick for intercepts: y-intercept is always easy—just let x = 0 and calculate! For x-intercepts, set y = 0 and solve. And remember: intercepts are points with two coordinates, so write them as (x, y), not just single numbers. When finding features of quadratics, you have two main forms: vertex form y = a(x - h)² + k shows the vertex directly, while standard form y = ax² + bx + c shows the y-intercept directly as (0, c). Choose your approach based on what form you're given!
The quadratic function g(x)=(x−1)2−9 is graphed on a coordinate plane. What is the vertex of the quadratic function g(x)=(x−1)2−9?
Explanation: This question tests your understanding of how to identify key features of quadratic functions and their graphs, specifically the vertex. For a quadratic function in the form y = a(x - h)² + k, the vertex is the point (h, k), which is the highest point if the parabola opens down (a < 0) or the lowest point if it opens up (a > 0). This quadratic is in vertex form y = a(x - h)² + k, where we can read the vertex directly: comparing g(x) = (x - 1)² - 9 to the standard form, we have a = 1, h = 1, and k = -9, so the vertex is (1, -9). The vertex is the turning point—the very top or very bottom of the parabola! Choice C is correct because it properly identifies the vertex as (1, -9) by reading directly from the vertex form. Great job identifying this! Choice A gives (1, 9) instead of (1, -9), making a sign error with the k-value. In the form (x - 1)² - 9, the -9 is the k-value, not +9. We all make sign errors—that's why checking our work is so important! With quadratics in vertex form y = a(x - h)² + k, the vertex is right in front of you: it's (h, k). Watch out though—there's a minus sign in the form, so if you see (x - 3)², the h-value is positive 3, but if you see (x + 3)², the h-value is -3!
A parabola is defined by h(x)=−(x−4)2+1. What is the maximum value of h(x)?
Explanation: This question tests your understanding of how to identify key features of quadratic functions and their graphs, specifically the maximum value. For a quadratic function in the form y = a(x - h)² + k, the vertex is the point (h, k), which is the highest point if the parabola opens down (a < 0) or the lowest point if it opens up (a > 0). This quadratic is in vertex form y = a(x - h)² + k, where we can read the vertex directly: h(x) = -(x - 4)² + 1 has vertex (4, 1). Since a = -1 is negative, the parabola opens down, making the vertex the maximum point. The maximum value is the y-coordinate of the vertex, which is 1. Choice C is correct because it properly identifies the maximum value as 1, which is the y-coordinate of the vertex (4, 1). Great job identifying this! Choice B gives 4, which is the x-coordinate of the vertex, not the maximum value. The maximum value is always the y-coordinate of the highest point, not the x-coordinate. We need the height, not the horizontal position! To remember parabola direction: positive a is a 'happy face' parabola opening up with a minimum, negative a is a 'sad face' opening down with a maximum. This visual trick helps you remember which is which!
A line is given by y=−2x+5. On a coordinate plane, what is the slope of this line?
Explanation: This question tests your understanding of how to identify key features of linear functions and their graphs, specifically slope. In a linear function y=mx+b, the slope m tells you how steep the line is and whether it rises (m>0) or falls (m<0), while the y-intercept b tells you where the line crosses the y-axis at the point (0,b). In the equation y=−2x+5, the slope is -2; this means for every 1 unit we move to the right, the graph goes down by 2 units, and a negative slope means the line falls from left to right. Choice B is correct because it properly identifies the slope as -2 by reading it directly from the coefficient of x in the slope-intercept form. Choice A confuses slope with y-intercept—an easy mix-up when you're learning! In y=mx+b, the number in front of x is the slope (m=−2), and the number by itself is the y-intercept (b=5). For linear functions in the form y=mx+b, you can read two key features immediately: m is the slope (how steep the line is), and b is the y-intercept (where it crosses the y-axis at (0,b)). No calculation needed—they're right there in the equation!
For the quadratic function y=−x2+4x−3, does the parabola open up or down?
Explanation: This question tests your understanding of how to identify key features of quadratic functions and their graphs, specifically direction of opening. A parabola's direction is determined by the sign of a (the coefficient of x²): if a is positive, the parabola opens upward like a smile (has a minimum), and if a is negative, it opens downward like a frown (has a maximum). Looking at the coefficient of x², which is -1, we can tell the direction: since a is negative, the parabola opens down; think of it this way: negative coefficient makes a 'sad' downward parabola! Choice B is correct because it properly identifies the direction as opens down based on the negative coefficient of x². Choice A gets the direction wrong: when the coefficient of x² is negative (like -1), the parabola opens downward and has a maximum, not a minimum or upward. To remember parabola direction: positive a is a 'happy face' parabola opening up with a minimum, negative a is a 'sad face' opening down with a maximum. This visual trick helps you remember which is which!
On a coordinate plane, graph y=x2−9. What are the x-intercepts (as points)?
Explanation: This question tests your understanding of how to identify key features of quadratic functions and their graphs, specifically x-intercepts. The x-intercept is where the graph crosses the x-axis, which always happens when y = 0: to find it, set your function equal to 0 and solve for x, giving you the point(s) (x-value, 0). To find the x-intercept(s), we set y = 0 and solve: 0 = x² - 9; this is a difference of squares, (x - 3)(x + 3) = 0, so x = 3 and x = -3, giving us the points (3, 0) and (-3, 0). Choice B is correct because it properly identifies the x-intercepts as (-3, 0) and (3, 0) using factoring, with accurate solutions. Choice A makes an error by listing y-intercepts instead: when you set y = 0 for x-intercepts, you solve for x, not y, so you get x-values ±3, not y-values ±9. Quick trick for intercepts: for x-intercepts, set y = 0 and solve. And remember: intercepts are points with two coordinates, so write them as (x, y), not just single numbers.
A line is modeled by y=21x−4. At what point does the graph cross the y-axis?
Explanation: This question tests your understanding of how to identify key features of linear functions and their graphs, specifically the y-intercept. The y-intercept is where the graph crosses the y-axis, which always happens when x=0: to find it, substitute x=0 into your function and you'll get the point (0,y-value). To find the y-intercept of y=21x−4, we substitute x=0 into the equation: y=21(0)−4=−4; this means the graph crosses the y-axis at the point (0,−4); it's the easiest intercept to find—just plug in 0 for x! Choice A is correct because it properly identifies the y-intercept as (0,−4) using substitution, with accurate calculation. Choice B mixes up the coordinates: intercepts are points, so we write them as (x,y) pairs; this choice has them backwards as (−4,0) instead of (0,−4); the x-value always comes first! Quick trick for intercepts: y-intercept is always easy—just let x=0 and calculate! And remember: intercepts are points with two coordinates, so write them as (x,y), not just single numbers.
A line is graphed on a coordinate plane using the equation y=2x−5. At what point does the graph cross the y-axis?
Explanation: This question tests your understanding of how to identify key features of linear functions and their graphs, specifically the y-intercept. The y-intercept is where the graph crosses the y-axis, which always happens when x = 0: to find it, substitute x = 0 into your function and you'll get the point (0, [y-value]). To find the y-intercept of y = 2x - 5, we substitute x = 0 into the equation: y = 2(0) - 5 = 0 - 5 = -5. This means the graph crosses the y-axis at the point (0, -5). It's the easiest intercept to find—just plug in 0 for x! Choice A is correct because it properly identifies the y-intercept as (0, -5) using substitution, with accurate calculation. Great job identifying this! Choice B mixes up the coordinates: intercepts are points, so we write them as (x, y) pairs. This choice has them backwards as (-5, 0) instead of (0, -5). The x-value always comes first! Quick trick for intercepts: y-intercept is always easy—just let x = 0 and calculate! For x-intercepts, set y = 0 and solve. And remember: intercepts are points with two coordinates, so write them as (x, y), not just single numbers.
For the linear function y=−x−2, at what point does the graph cross the y-axis?
Explanation: This question tests your understanding of how to identify key features of linear functions and their graphs, specifically the y-intercept. The y-intercept is where the graph crosses the y-axis, which always happens when x = 0: to find it, substitute x = 0 into your function and you'll get the point (0, y-value). To find the y-intercept of y = -x - 2, we substitute x = 0 into the equation: y = -(0) - 2 = -2; this means the graph crosses the y-axis at the point (0, -2); it's the easiest intercept to find—just plug in 0 for x! Choice B is correct because it properly identifies the y-intercept as (0, -2) using substitution, with accurate calculation. Choice A mixes up the coordinates: intercepts are points, so we write them as (x, y) pairs; this choice has (-2, 0) which is actually the x-intercept, not the y-intercept. Quick trick for intercepts: y-intercept is always easy—just let x = 0 and calculate! And remember: intercepts are points with two coordinates, so write them as (x, y), not just single numbers.
The quadratic function y=−(x−3)2+4 is graphed on a coordinate plane. What is the maximum value of the function?
Explanation: This question tests your understanding of how to identify key features of quadratic functions and their graphs, specifically the vertex and its relation to maximum value. For a quadratic function in the form y = a(x - h)² + k, the vertex is the point (h, k), which is the highest point if the parabola opens down (a < 0) or the lowest point if it opens up (a > 0). This quadratic is in vertex form y = -1(x - 3)² + 4, where we can read the vertex directly: it's (h, k) = (3, 4); since a < 0, the maximum value is the y-coordinate of the vertex, which is 4—the turning point is the very top of the parabola! Choice C is correct because it properly identifies the maximum value as 4 by reading the k value from vertex form, since the parabola opens down. Great job identifying this! Choice A gets the sign wrong: when the coefficient of the squared term is negative (like -1), the parabola opens downward and has a maximum of 4, not -4—think: negative coefficient = sad face = opens down with max at top! With quadratics in vertex form y = a(x - h)² + k, the vertex is right in front of you: it's (h, k). Watch out though—there's a minus sign in the form, so if you see (x - 3)², the h-value is positive 3, but if you see (x + 3)², the h-value is -3!
Find both intercepts of the linear function y=2x−4. Give your answer as points.
Explanation: This question tests your understanding of how to identify key features of linear functions and their graphs, specifically intercepts. The y-intercept is where the graph crosses the y-axis, which always happens when x = 0: to find it, substitute x = 0 into your function and you'll get the point (0, [y-value]). The x-intercept is where the graph crosses the x-axis, which always happens when y = 0: to find it, set your function equal to 0 and solve for x, giving you the point(s) ([x-value], 0). To find the y-intercept of y = 2x - 4, we substitute x = 0 into the equation: y = 2(0) - 4 = 0 - 4 = -4. This means the graph crosses the y-axis at the point (0, -4). To find the x-intercept, we set y = 0 and solve: 0 = 2x - 4. Solving gives 2x = 4, so x = 2, and the x-intercept is (2, 0). The intercepts show where the graph touches the axes. Choice A is correct because it properly identifies both intercepts as (2, 0) for the x-intercept and (0, -4) for the y-intercept, with accurate calculations. Great job identifying this! Choice B mixes up the coordinates: intercepts are points, so we write them as (x, y) pairs. This choice has them backwards, putting the x-intercept at (0, -4) and y-intercept at (2, 0) instead of the correct way around. The x-value always comes first! Quick trick for intercepts: y-intercept is always easy—just let x = 0 and calculate! For x-intercepts, set y = 0 and solve. And remember: intercepts are points with two coordinates, so write them as (x, y), not just single numbers.
For the quadratic function y=(x−1)2−5, does the parabola open up or down?
Explanation: This question tests your understanding of how to identify key features of quadratic functions and their graphs, specifically direction of opening. A parabola's direction is determined by the sign of a (the coefficient of x2): if a is positive, the parabola opens upward like a smile (has a minimum), and if a is negative, it opens downward like a frown (has a maximum). This quadratic is y=(x−1)2−5, which expands to y=x2−2x−4, so the coefficient of x2 is 1, and since a is positive, the parabola opens up; think of it this way: positive coefficient makes a 'happy' upward parabola! Choice B is correct because it properly identifies the direction as opens up based on the positive coefficient of x2. Choice A gets the direction wrong: when the coefficient of x2 is positive (like 1), the parabola opens upward and has a minimum, not downward. To remember parabola direction: positive a is a 'happy face' parabola opening up with a minimum, negative a is a 'sad face' opening down with a maximum. This visual trick helps you remember which is which!
On a coordinate plane, the quadratic function y=x2−9 is graphed. What are the x-intercepts of y=x2−9?
Explanation: This question tests your understanding of how to identify key features of quadratic functions and their graphs, specifically the x-intercepts. The x-intercept is where the graph crosses the x-axis, which always happens when y = 0: to find it, set your function equal to 0 and solve for x, giving you the point(s) ([x-value], 0). To find the x-intercept(s), we set y = 0 and solve: 0 = x² - 9. Factoring gives us (x - 3)(x + 3) = 0, so x = 3 and x = -3, giving us the points (-3, 0) and (3, 0). The x-intercepts show where the graph touches or crosses the x-axis. Choice C is correct because it properly identifies the x-intercepts as (-3, 0) and (3, 0) using the difference of squares factoring, with accurate calculation. Great job identifying this! Choice D gives just the numbers -3 and 3 instead of the points (-3, 0) and (3, 0). Remember: intercepts are points on the graph, not just single numbers, so we need both coordinates! Quick trick for intercepts: y-intercept is always easy—just let x = 0 and calculate! For x-intercepts, set y = 0 and solve. And remember: intercepts are points with two coordinates, so write them as (x, y), not just single numbers.
At what point does the graph of y=21x−3 cross the y-axis?
Explanation: This question tests your understanding of how to identify key features of linear functions and their graphs, specifically the y-intercept. The y-intercept is where the graph crosses the y-axis, which always happens when x = 0: to find it, substitute x = 0 into your function and you'll get the point (0, [y-value]). To find the y-intercept of y = ½x - 3, we substitute x = 0 into the equation: y = ½(0) - 3 = 0 - 3 = -3. This means the graph crosses the y-axis at the point (0, -3). It's the easiest intercept to find—just plug in 0 for x! Choice A is correct because it properly identifies the y-intercept as (0, -3) using direct substitution, with accurate calculation. Great job identifying this! Choice D gives just the number -3 instead of the point (0, -3). Remember: intercepts are points on the graph, not just single numbers, so we need both coordinates! Quick trick for intercepts: y-intercept is always easy—just let x = 0 and calculate! For x-intercepts, set y = 0 and solve. And remember: intercepts are points with two coordinates, so write them as (x, y), not just single numbers.
For the quadratic function y=(x−4)2, what is the vertex of the parabola on the coordinate plane?
Explanation: This question tests your understanding of how to identify key features of quadratic functions and their graphs, specifically the vertex. For a quadratic function in the form y=a(x−h)2+k, the vertex is the point (h,k), which is the highest point if the parabola opens down (a<0) or the lowest point if it opens up (a>0). This quadratic is in vertex form y=1(x−4)2+0, where we can read the vertex directly: it's (h,k)=(4,0); the vertex is the turning point—the very bottom of the parabola since a>0! Choice B is correct because it properly identifies the vertex as (4,0) by reading it from the vertex form, with accurate values. Choice A mixes up the coordinates: the vertex is (4,0), not (0,4); remember, h is the x-coordinate from (x−h), and k is the y-coordinate. With quadratics in vertex form y=a(x−h)2+k, the vertex is right in front of you: it's (h,k). Watch out though—there's a minus sign in the form, so if you see (x−3)2, the h-value is positive 3, but if you see (x+3)2, the h-value is -3!