On a coordinate plane, graph the linear function . Find both intercepts of the function (as points).
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Algebra · Learn by Concept
Review real example questions for Graph Linear And Quadratic Functions in Algebra.
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On a coordinate plane, graph the linear function y=−x+2. Find both intercepts of the function (as points).
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!