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Algebra Help: Graph Linear And Quadratic Functions

Review real example questions for Graph Linear And Quadratic Functions in Algebra.

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A line is given by $y=-2x+5$ . On a coordinate plane, what is the slope of this line?

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Question 1

A line is given by $y=-2x+5$ . On a coordinate plane, what is the slope of this line?

$5$
$-5$
$-2$ (correct answer)
$2$

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!

Question 2

On a coordinate plane, the quadratic function $y = x^2 - 5x + 6$ is graphed. What are the x-intercepts of $y = x^2 - 5x + 6$ (as points)?

$(5, 0)$ and $(6, 0)$
$(-2, 0)$ and $(-3, 0)$
$(2, 0)$ and $(3, 0)$ (correct answer)
$(0, 2)$ and $(0, 3)$

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!

Question 3

For the linear function $y=-x-2$ , at what point does the graph cross the $y$ -axis?

$(0,-2)$ (correct answer)
$(2,0)$
$(0,2)$
$(-2,0)$

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.

Question 4

At what point does the graph of $y = \frac{1}{2}x - 3$ cross the y-axis?

$(0, -3)$ (correct answer)
$(-3, 0)$
$(0, 3)$
$-3$

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.

Question 5

A parabola is graphed on a coordinate plane for the function $f(x) = (x - 1)^2 - 4$ . What is the vertex of the quadratic function $f(x) = (x - 1)^2 - 4$ ?

$(1, -4)$ (correct answer)
$(-1, -4)$
$(1, 4)$
$(-4, 1)$

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: it's (h, k) = (1, -4). The vertex is the turning point—the very top or very bottom of the parabola! Choice A is correct because it properly identifies the vertex as (1, -4) using the vertex form, with accurate reading of h and k values. Great job identifying this! Choice B gives (-1, -4), which makes an error with the sign of h. Watch out though—there's a minus sign in the form, so if you see (x - 1)², the h-value is positive 1, not negative 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!

Question 6

A line is graphed on a coordinate plane and crosses the $y$ -axis at $-7$ . The line is described by the equation $y = 4x - 7$ . At what point does the graph cross the $y$ -axis?

$(7, 0)$
$(-7, 0)$
$(0, 7)$
$(0, -7)$ (correct answer)

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 = 4x - 7, we substitute x = 0 into the equation: y = 4(0) - 7 = -7. This means the graph crosses the y-axis at the point (0, -7). 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, -7) by substituting x = 0, with accurate calculation. Great job identifying this! Choice A gives just the number -7 as ( -7, 0 ) instead of (0, -7). 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.

Question 7

On a coordinate plane, graph the linear function $y = -x + 2$ . Find both intercepts of the function (as points).

$(2, 0)$ and $(0, -2)$
$(0, 1)$ and $(1, 0)$
$(0, -2)$ and $(-2, 0)$
$(0, 2)$ and $(2, 0)$ (correct answer)

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!

Question 8

A line is modeled by $y=\frac{1}{2}x-4$ . At what point does the graph cross the $y$ -axis?

$(-4,0)$
$(4,0)$
$(0,-4)$ (correct answer)
$(0,4)$

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 $(0, \text{y-value})$ . To find the y-intercept of $y = \frac{1}{2}x - 4$ , we substitute $x = 0$ into the equation: $y = \frac{1}{2}(0) - 4 = -4$ ; 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.

Question 9

A line is graphed on a coordinate plane using the equation $y = 2x - 5$ . At what point does the graph cross the y-axis?

$(0, -5)$ (correct answer)
$(-5, 0)$
$(0, 5)$
$-5$

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.

Question 10

Graph the quadratic function $y=(x+1)^2+2$ on a coordinate plane. What is the vertex?

$(-1,-2)$
$(2,-1)$
$(1,2)$
$(-1,2)$ (correct answer)

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!