Graph Linear and Quadratic Functions - Algebra 1

Graph the quadratic function $f(x)=(x-2)^2-3$ on a coordinate plane. What is the vertex?
Answer: $(2,-3)$ 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 $y = 1(x - 2)^2 + (-3)$ , where we can read the vertex directly: it's $(h, k) = (2, -3)$ ; the vertex is the turning point—the very bottom of the parabola since $a > 0$ ! Choice A is correct because it properly identifies the vertex as $(2, -3)$ by reading it from the vertex form, with accurate values. Choice B makes an error in the vertex calculation, using the wrong sign for h: the form is $(x - h)$ , so $(x - 2)$ means $h = 2$ , not $-2$ . 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 $-3$ !
The quadratic function $y = - (x - 3)^2 + 4$ is graphed on a coordinate plane. What is the maximum value of the function?
Answer: $4$ 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!
For the quadratic function $y=-2x^2+8x-6$ , what is the maximum value of the function?
Answer: $2$ This question tests your understanding of how to identify key features of quadratic functions and their graphs, specifically the vertex and its y-value as the maximum. 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 $a = -2 < 0$ , the parabola opens down and has a maximum at the vertex; using the formula $x = -\frac{b}{2a} = -\frac{8}{2(-2)} = -\frac{8}{-4} = 2$ , then $y = -2(2)^2 + 8(2) - 6 = -8 + 16 - 6 = 2$ , so the maximum is 2. Choice A is correct because it properly identifies the maximum value as 2 using the vertex formula, with accurate calculation. Choice B makes an arithmetic mistake when calculating y, computing -6 instead of 2; we all make arithmetic errors—that's why checking our work is so important! 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!
Graph the quadratic function $f(x) = -x^2 + 4x$ on a coordinate plane. What is the vertex of the parabola?
Answer: $(2, 4)$ 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 ( $a > 0$ ). Using the formula $x = -b/(2a)$ for standard form $y = -x^2 + 4x$ , we find $x = -4/(2 \cdot -1) = -4/-2 = 2$ , and substituting back gives $y = -(2)^2 + 4(2) = -4 + 8 = 4$ , so vertex is $(2, 4)$ . 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 $(2, 4)$ using the vertex formula, with accurate calculation. Great job identifying this! Choice D mixes up the coordinates: intercepts are points, so we write them as $(x, y)$ pairs. This choice has them backwards as $(4, 2)$ instead of $(2, 4)$ . The x-value always comes first! 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 $y = ax^2 + bx + c$ shows the y-intercept directly as $(0, c)$ . Choose your approach based on what form you're given!
A linear function is given by $y=\tfrac{1}{2}x+4$ . What is the y-intercept of $y=\tfrac{1}{2}x+4$ ? (Give your answer as a point.)
Answer: $(0,4)$ 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 + 4, we substitute x = 0 into the equation: y = ½(0) + 4 = 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 B is correct because it properly identifies the y-intercept as (0, 4) using direct substitution, with accurate calculation. Great job identifying this! Choice C gives just the number 4 instead of the point (0, 4). Remember: intercepts are points on the graph, not just single numbers, so we need both coordinates! 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!
What is the maximum value of the function $h(t) = -t^2 + 4t + 1$ ?
Answer: $5$ 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!
Consider the quadratic function $y=-2(x+3)^2+5$ on a coordinate plane. Does the parabola $y=-2(x+3)^2+5$ open up or down?
Answer: Opens down This question tests your understanding of how to identify key features of quadratic functions and their graphs, specifically the 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 -2 in the equation y = -2(x + 3)² + 5, we can tell the direction: since a is negative, the parabola opens down. Think of it this way: positive coefficient makes a 'happy' upward parabola, negative makes a 'sad' downward one! Choice B is correct because it properly identifies that the parabola opens down due to the negative coefficient a = -2. Great job identifying this! Choice A gets the direction wrong: when the coefficient of x² is negative (like -2), the parabola opens downward and has a maximum, not upward with a minimum. Think: negative coefficient = sad face = opens down! 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 the linear function $y = 3x - 6$ . Find both intercepts of the function (as points).
Answer: $(0, -6)$ and $(2, 0)$ 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 = 3x - 6, we substitute x = 0 into the equation: y = 3(0) - 6 = -6, so the graph crosses the y-axis at (0, -6); for the x-intercept, we set y = 0 and solve 0 = 3x - 6, which gives 3x = 6, x = 2, so the point is (2, 0). Choice B is correct because it properly identifies the intercepts as (0, -6) and (2, 0) using substitution and solving, with accurate calculations. Great job identifying this! Choice A makes an arithmetic mistake when finding the y-intercept, computing 3(0) - 6 as +6 instead of -6—we all make sign errors sometimes, so double-check those operations! 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 function $y=-3x+9$ , find both intercepts of the line (as points) to show on the coordinate plane.
Answer: $(0,9)$ and $(3,0)$ 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 = -3x + 9, we substitute x = 0: y = -3(0) + 9 = 9, so (0, 9); for the x-intercept, set y = 0: 0 = -3x + 9, so 3x = 9 and x = 3, giving (3, 0). Choice B is correct because it properly identifies the intercepts as (0, 9) and (3, 0) using substitution and solving, with accurate calculations. Choice D makes an arithmetic mistake when finding the y-intercept, computing y = 3 instead of 9; we all make arithmetic 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.
A line is graphed on a coordinate plane and is represented by the equation $y = -2x + 5$ . For the function $y = -2x + 5$ , what is the slope?
Answer: $-2$ 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—a negative slope means the line falls from left to right. Choice A is correct because it properly identifies the slope as -2 by reading the coefficient of x in the equation, with no calculation needed. Great job identifying this! Choice B confuses the sign of the slope—an easy mix-up when you're learning! In y = mx + b, the number in front of x is the slope (m = -2), including the negative sign, not just the positive value. 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 $y = 4x - 8$ , find both intercepts of the function.
Answer: x-intercept $(2, 0)$ and y-intercept $(0, -8)$ This question tests your understanding of how to identify key features of linear functions and their graphs, specifically both 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]). To find the y-intercept of y = 4x - 8, we substitute x = 0: y = 4(0) - 8 = -8, giving us (0, -8). To find the x-intercept, we set y = 0 and solve: 0 = 4x - 8. Solving gives 4x = 8, so x = 2, and the x-intercept is (2, 0). The x-intercepts show where the graph touches or crosses the x-axis. Choice B is correct because it properly identifies both intercepts as (2, 0) for the x-intercept and (0, -8) for the y-intercept, with accurate calculations. Great job identifying this! Choice C gives just the numbers 2 and -8 instead of the points (2, 0) and (0, -8). 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.
A parabola is graphed on a coordinate plane and is represented by $y = x^2 + 6x + 8$ . What is the vertex of the quadratic function $y = x^2 + 6x + 8$ ?
Answer: $(-3, -1)$ 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!
For the function $y=2x+7$ , find both intercepts of the line (as points) to label on the graph.
Answer: $(0,7)$ and $\left(-\frac{7}{2},0\right)$ 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 = 2x + 7, we substitute x = 0: y = 2(0) + 7 = 7, so (0, 7); for the x-intercept, set y = 0: 0 = 2x + 7, so 2x = -7 and x = -7/2, giving (-7/2, 0). Choice B is correct because it properly identifies the intercepts as (0, 7) and (-7/2, 0) using substitution and solving, with accurate calculations. Choice D makes an arithmetic mistake when finding the y-intercept, computing y = -7 instead of 7; remember to include the sign correctly when solving! 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.
A student graphs the function $y = 3x + 4$ on a coordinate plane. What is the y-intercept of $y = 3x + 4$ ?
Answer: $(0, 4)$ 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 = 3x + 4, we substitute x = 0 into the equation: y = 3(0) + 4 = 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 B is correct because it properly identifies the y-intercept as (0, 4) using the standard form y = mx + b, where b = 4 is the y-intercept. Great job identifying this! Choice C gives just the number 4 instead of the point (0, 4). Remember: intercepts are points on the graph, not just single numbers, so we need both coordinates! 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!
What are the x-intercepts of the quadratic function $y = x^2 + x - 6$ ?
Answer: $(2, 0)$ and $(-3, 0)$ 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 + x - 6. Factoring gives us 0 = (x + 3)(x - 2), so x = -3 and x = 2, giving us the points (-3, 0) and (2, 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) using factoring, with accurate calculation. Great job identifying this! Choice B gives (-2, 0) and (3, 0), which reverses the signs of the correct x-values. When factoring (x + 3)(x - 2) = 0, we get x = -3 (not 3) and x = 2 (not -2). We all make sign errors—that's why checking our work is so important! 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!
A parabola is defined by $h(x)=-(x-4)^2+1$ . What is the maximum value of $h(x)$ ?
Answer: $1$ 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!
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)?
Answer: $(2, 0)$ and $(3, 0)$ 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!
Find both intercepts of the linear function $y = 2x - 4$ . Give your answer as points.
Answer: x-intercept $(2, 0)$ and y-intercept $(0, -4)$ 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.
On a coordinate plane, graph $y=(x+2)(x-1)$ . What are the $x$ -intercepts (as points)?
Answer: $(-2,0)$ and $(1,0)$ 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). Since it's already factored as y = (x + 2)(x - 1), set 0 = (x + 2)(x - 1), so x + 2 = 0 or x - 1 = 0, giving x = -2 and x = 1, and points (-2, 0) and (1, 0). Choice B is correct because it properly identifies the x-intercepts as (-2, 0) and (1, 0) using the factored form, with accurate solutions. Choice A gets the signs wrong: from (x + 2) = 0 gives x = -2, and (x - 1) = 0 gives x = 1, not positive 2 and negative 1. 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 parabola is graphed for the function $y = -x^2 + 4x + 1$ on a coordinate plane. What is the maximum value of $y = -x^2 + 4x + 1$ ?
Answer: $5$ This question tests your understanding of how to identify key features of quadratic functions and their graphs, specifically the maximum value. 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). Since the coefficient of x² is -1 (negative), this parabola opens down and has a maximum. Using the formula x = -b/(2a), we find x = -(4)/(2·(-1)) = -4/-2 = 2, and substituting back gives y = -(2)² + 4(2) + 1 = -4 + 8 + 1 = 5, so the vertex is (2, 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, which occurs at the vertex (2, 5). Great job identifying this! Choice B gives 1, which is actually the y-intercept (when x = 0, y = 1), not the maximum. The maximum occurs at the vertex, not at the y-intercept! 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). For maximum/minimum in standard form, use x = -b/(2a) to find the vertex!

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