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!

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!

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!

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!

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!

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!

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!

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.

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 $x^2$): 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 $y = x^2 - 2x - 4$, so the coefficient of $x^2$ 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 $x^2$. Choice A gets the direction wrong: when the coefficient of $x^2$ 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!

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.