Algebra Quiz: Graph Linear And Quadratic Functions
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Graph Linear And Quadratic FunctionsQuestion 1 of 20

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

55
−5-5
−2-2
22
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Algebra Quiz

Algebra Quiz: Graph Linear And Quadratic Functions

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.

What this quiz covers

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.

How to use this quiz

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.

All questions

Question 1

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

  1. 55
  2. −5-5
  3. −2-2 (correct answer)
  4. 22
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+by = mx + b, the slope mm tells you how steep the line is and whether it rises (m>0m > 0) or falls (m<0m < 0), while the y-intercept bb tells you where the line crosses the y-axis at the point ($0, b$). In the equation y=−2x+5y = -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+by = mx + b, the number in front of x is the slope (m=−2m = -2), and the number by itself is the y-intercept (b=5b = 5). For linear functions in the form y=mx+by = mx + b, you can read two key features immediately: mm is the slope (how steep the line is), and bb 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=x2−5x+6y = x^2 - 5x + 6 is graphed. What are the x-intercepts of y=x2−5x+6y = x^2 - 5x + 6 (as points)?

  1. (5,0)(5, 0) and (6,0)(6, 0)
  2. (−2,0)(-2, 0) and (−3,0)(-3, 0)
  3. (2,0)(2, 0) and (3,0)(3, 0) (correct answer)
  4. (0,2)(0, 2) and (0,3)(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−2y=-x-2, at what point does the graph cross the yy-axis?

  1. (0,−2)(0,-2) (correct answer)
  2. (2,0)(2,0)
  3. (0,2)(0,2)
  4. (−2,0)(-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=12x−3y = \frac{1}{2}x - 3 cross the y-axis?​

  1. (0,−3)(0, -3) (correct answer)
  2. (−3,0)(-3, 0)
  3. (0,3)(0, 3)
  4. −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)2−4f(x) = (x - 1)^2 - 4. What is the vertex of the quadratic function f(x)=(x−1)2−4f(x) = (x - 1)^2 - 4?

  1. (1,−4)(1, -4) (correct answer)
  2. (−1,−4)(-1, -4)
  3. (1,4)(1, 4)
  4. (−4,1)(-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 yy-axis at −7-7. The line is described by the equation y=4x−7y = 4x - 7. At what point does the graph cross the yy-axis?

  1. (7,0)(7, 0)
  2. (−7,0)(-7, 0)
  3. (0,7)(0, 7)
  4. (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+2y = -x + 2. Find both intercepts of the function (as points).

  1. (2,0)(2, 0) and (0,−2)(0, -2)
  2. (0,1)(0, 1) and (1,0)(1, 0)
  3. (0,−2)(0, -2) and (−2,0)(-2, 0)
  4. (0,2)(0, 2) and (2,0)(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=12x−4y=\frac{1}{2}x-4. At what point does the graph cross the yy-axis?

  1. (−4,0)(-4,0)
  2. (4,0)(4,0)
  3. (0,−4)(0,-4) (correct answer)
  4. (0,4)(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=0x = 0: to find it, substitute x=0x = 0 into your function and you'll get the point (0,y-value)(0, \text{y-value}). To find the y-intercept of y=12x−4y = \frac{1}{2}x - 4, we substitute x=0x = 0 into the equation: y=12(0)−4=−4y = \frac{1}{2}(0) - 4 = -4; this means the graph crosses the y-axis at the point (0,−4)(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)(0, -4) using substitution, with accurate calculation. Choice B mixes up the coordinates: intercepts are points, so we write them as (x,y)(x, y) pairs; this choice has them backwards as (−4,0)(-4, 0) instead of (0,−4)(0, -4); the x-value always comes first! Quick trick for intercepts: y-intercept is always easy—just let x=0x = 0 and calculate! And remember: intercepts are points with two coordinates, so write them as (x,y)(x, y), not just single numbers.

Question 9

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

  1. (0,−5)(0, -5) (correct answer)
  2. (−5,0)(-5, 0)
  3. (0,5)(0, 5)
  4. −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+2y=(x+1)^2+2 on a coordinate plane. What is the vertex?

  1. (−1,−2)(-1,-2)
  2. (2,−1)(2,-1)
  3. (1,2)(1,2)
  4. (−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!

Question 11

The quadratic function is y=x2+6x+5y=x^2+6x+5. What is the y-intercept of y=x2+6x+5y=x^2+6x+5? (Give your answer as a point.)

  1. (5,0)(5,0)
  2. 55
  3. (0,5)(0,5) (correct answer)
  4. (0,−5)(0,-5)
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 = x² + 6x + 5, we substitute x = 0 into the equation: y = (0)² + 6(0) + 5 = 0 + 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 B is correct because it properly identifies the y-intercept as (0, 5) using direct substitution, with accurate calculation. Great job identifying this! Choice D gives just the number 5 instead of the point (0, 5). Remember: intercepts are points on the graph, not just single numbers, so we need both coordinates! 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 12

For the quadratic function y=2x2−8x+6y=2x^2-8x+6, what is the yy-intercept (as a point) to label on the graph?

  1. (6,0)(6,0)
  2. (−6,0)(-6,0)
  3. (0,2)(0,2)
  4. (0,6)(0,6) (correct answer)
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.

Question 13

A parabola is graphed on a coordinate plane and is represented by y=x2+6x+8y = x^2 + 6x + 8. What is the vertex of the quadratic function y=x2+6x+8y = x^2 + 6x + 8?

  1. (3,−1)(3, -1)
  2. (−3,−1)(-3, -1) (correct answer)
  3. (3,17)(3, 17)
  4. (−3,1)(-3, 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). Since this quadratic is in standard form y = x² + 6x + 8, we use the formula x = -b/(2a) to find the vertex: x = -6/(2·1) = -3, and substituting back gives y = (-3)² + 6(-3) + 8 = 9 - 18 + 8 = -1, so vertex is (-3, -1); 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 (-3, -1) using the vertex formula and substitution, with accurate calculation. Great job identifying this! Choice C makes an error in the vertex calculation, using the y-value as -1 but the x-value as positive 3 instead of -3; the formula x = -b/(2a) requires - (6)/(2) = -3, not +3. 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 14

The quadratic function y=−(x−3)2+4y = - (x - 3)^2 + 4 is graphed on a coordinate plane. What is the maximum value of the function?

  1. 44 (correct answer)
  2. −4-4
  3. 77
  4. 33
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!

Question 15

On a coordinate plane, graph the quadratic function y=x2−5x+6y = x^2 - 5x + 6. What are the xx-intercepts (as points)?

  1. (0,2)(0, 2) and (0,3)(0, 3)
  2. (2,0)(2, 0) and (3,0)(3, 0) (correct answer)
  3. (−2,0)(-2, 0) and (−3,0)(-3, 0)
  4. (6,0)(6, 0) and (1,0)(1, 0)
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.

Question 16

Find both intercepts of the linear function y=2x−4y = 2x - 4. Give your answer as points.

  1. x-intercept (2,0)(2, 0) and y-intercept (0,−4)(0, -4) (correct answer)
  2. x-intercept (0,−4)(0, -4) and y-intercept (2,0)(2, 0)
  3. x-intercept 22 and y-intercept −4-4
  4. x-intercept (−2,0)(-2, 0) and y-intercept (0,4)(0, 4)
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.

Question 17

For the function y=4x+1y=4x+1, find both intercepts of the line (as points) to show on the coordinate plane.

  1. (0,1)(0,1) and (−14,0)\left(-\frac{1}{4},0\right) (correct answer)
  2. (1,0)(1,0) and (0,−14)\left(0,-\frac{1}{4}\right)
  3. (0,−1)(0,-1) and (14,0)\left(\frac{1}{4},0\right)
  4. (0,4)(0,4) and (−1,0)\left(-1,0\right)
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 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, b). To find the y-intercept of y = 4x + 1, we substitute x = 0: y = 4(0) + 1 = 1, so (0, 1); for the x-intercept, set y = 0: 0 = 4x + 1, so 4x = -1 and x = -1/4, giving (-1/4, 0). Choice A is correct because it properly identifies the intercepts as (0, 1) and (-1/4, 0) using substitution and solving, with accurate calculations. Choice B mixes up the coordinates: intercepts are points, so we write them as (x, y) pairs, but this choice has them backwards like (1, 0) instead of (0, 1). 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 18

On a coordinate plane, the quadratic function y=x2−9y = x^2 - 9 is graphed. What are the x-intercepts of y=x2−9y = x^2 - 9?

  1. (−9,0)(-9, 0) and (9,0)(9, 0)
  2. (0,−9)(0, -9) and (0,9)(0, 9)
  3. (−3,0)(-3, 0) and (3,0)(3, 0) (correct answer)
  4. −3-3 and 33
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.

Question 19

What is the maximum value of the function h(t)=−t2+4t+1h(t) = -t^2 + 4t + 1?​

  1. 11
  2. 44
  3. 55 (correct answer)
  4. −5-5
Explanation: This question tests your understanding of how to identify key features of quadratic functions and their graphs, specifically finding the maximum value. A parabola's direction is determined by the sign of a (the coefficient of t²): 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). Since the coefficient of t² is -1 (negative), this parabola opens down and has a maximum at its vertex. Using the formula t = -b/(2a), we find t = -4/(2·(-1)) = -4/(-2) = 2, and substituting back gives h(2) = -(2)² + 4(2) + 1 = -4 + 8 + 1 = 5, so the maximum value is 5. The vertex is the turning point—the very top or very bottom of the parabola! Choice C is correct because it properly identifies the maximum value as 5 by finding the vertex using the formula t = -b/(2a), with accurate calculation. Great job identifying this! Choice B gives 4, which might come from confusing the coefficient of t (which is 4) with the maximum value. The maximum occurs at the vertex, not from reading coefficients directly! 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 20

For the quadratic function y=(x−4)2y=(x-4)^2, what is the vertex of the parabola on the coordinate plane?

  1. (4,0)(4,0) (correct answer)
  2. (−4,0)(-4,0)
  3. (0,4)(0,4)
  4. (0,−4)(0,-4)
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+ky = a(x - h)^2 + k, the vertex is the point (h,k)(h, k), which is the highest point if the parabola opens down (a<0a < 0) or the lowest point if it opens up (a>0a > 0). This quadratic is in vertex form y=1(x−4)2+0y = 1(x - 4)^2 + 0, where we can read the vertex directly: it's (h,k)=(4,0)(h, k) = (4, 0); the vertex is the turning point—the very bottom of the parabola since a>0a > 0! Choice B is correct because it properly identifies the vertex as (4,0)(4, 0) by reading it from the vertex form, with accurate values. Choice A mixes up the coordinates: the vertex is (4,0)(4, 0), not (0,4)(0, 4); remember, h is the x-coordinate from (x−h)(x - h), and k is the y-coordinate. With quadratics in vertex form y=a(x−h)2+ky = a(x - h)^2 + k, the vertex is right in front of you: it's (h,k)(h, k). Watch out though—there's a minus sign in the form, so if you see (x−3)2(x - 3)^2, the h-value is positive 3, but if you see (x+3)2(x + 3)^2, the h-value is -3!