SAT Math Quiz: Graphing Functions

What this quiz covers

This quiz focuses on Graphing Functions, giving you a quick way to practice the rules, question types, and explanations that matter most for SAT Math.

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.

Question 1

1. $f(x)=2x-2$ (correct answer)
2. $f(x)=\frac{2}{3}x-2$
3. $f(x)=\frac{3}{2}x-2$
4. $f(x)=-2x-2$

Explanation: We need to find the equation of a line passing through $(0, -2)$ and $(3, 4)$. The y-intercept is already given as -2 (when $x = 0$, $y = -2$). Calculate the slope: $m = \frac{4 - (-2)}{3 - 0} = \frac{6}{3} = 2$. Using slope-intercept form: $f(x) = mx + b = 2x - 2$. To verify, check both points: $f(0) = 2(0) - 2 = -2$ ✓ and $f(3) = 2(3) - 2 = 6 - 2 = 4$ ✓. A common error is calculating slope as $\Delta x / \Delta y = 3/6 = 1/2$ instead of $\Delta y / \Delta x = 6/3 = 2$. When finding linear equations, always verify your answer by substituting both given points.

Question 2

1. Yes, for all $x$
2. No, it fails the vertical line test (correct answer)
3. Yes, except at $x=0$
4. Cannot be determined

Explanation: The graph shows a circle centered at the origin with radius 3. To determine if this represents a function of x, we apply the vertical line test: does any vertical line intersect the graph at more than one point? For any x-value between -3 and 3 (except x = -3 and x = 3), a vertical line will intersect the circle at two points - one on the upper semicircle and one on the lower semicircle. For example, at x = 0, the circle passes through both (0, 3) and (0, -3). Since vertical lines intersect the graph at multiple points, this fails the vertical line test and is not a function of x. The key insight is that circles (except for single points) always fail the vertical line test because they curve back on themselves.

Question 3

1. $f(x)=2^x$
2. $f(x)=4\left(\tfrac12\right)^x$ (correct answer)
3. $f(x)=4\cdot 2^x$
4. $f(x)=2\left(\tfrac12\right)^x$

Explanation: The exponential function passes through (0, 4) and (1, 2), showing decay since the y-values decrease as x increases. For exponential functions f(x) = a·b^x, the point (0, 4) gives us a = 4. Using point (1, 2): 2 = 4·b^1, so b = 1/2. Therefore, f(x) = 4(1/2)^x. This matches option B. To verify the decay: as x increases, (1/2)^x decreases since the base is between 0 and 1. A common mistake is confusing growth and decay functions, or using 2^x instead of (1/2)^x. When identifying exponential functions from graphs, always check whether y-values increase (growth) or decrease (decay) as x increases.

Question 4

1. $[-4,2]$ (correct answer)
2. $[-1,2]$
3. $(-4,2)$
4. $[-4,-1]$

Explanation: The graph shows a line segment from (-4, 2) to (2, -1) with solid endpoints. The domain of a function is the set of all possible x-values. Looking at the x-coordinates, the segment extends from x = -4 to x = 2. Since the endpoints are solid (filled in), both -4 and 2 are included in the domain. Therefore, the domain is [-4, 2], using square brackets to indicate inclusion of endpoints. A common mistake is giving the range [-1, 2] instead of the domain, or using parentheses (-4, 2) which would exclude the endpoints. When finding domain from a graph, focus on the horizontal extent and pay attention to whether endpoints are included (solid dots) or excluded (open circles).

Question 5

1. $f(x)=\dfrac{1}{x-1}+2$ (correct answer)
2. $f(x)=\dfrac{1}{x+1}+2$
3. $f(x)=\dfrac{1}{x-1}-2$
4. $f(x)=\dfrac{2}{x-1}+1$

Explanation: The rational function has a vertical asymptote at x = 1, horizontal asymptote at y = 2, and passes through (0, 1). A vertical asymptote at x = 1 means the denominator has factor (x - 1). A horizontal asymptote at y = 2 suggests the function has form f(x) = 1/(x - 1) + 2. Let's verify with point (0, 1): f(0) = 1/(0 - 1) + 2 = 1/(-1) + 2 = -1 + 2 = 1 ✓. This confirms option A is correct. Common errors include confusing vertical and horizontal asymptotes, or using the wrong sign in the denominator. When dealing with rational functions, always verify your equation using a known point on the graph.

Question 6

1. $g(x)=f(x-2)$
2. $g(x)=f(x)+2$ (correct answer)
3. $g(x)=f(x+2)$
4. $g(x)=-f(x)+2$

Explanation: We need to identify the transformation from f(x) with vertex at (0, 0) to g(x) with vertex at (0, 2), where both parabolas have the same shape. Since g(x) has the same shape as f(x) but its vertex is shifted from (0, 0) to (0, 2), this is a vertical shift up by 2 units. This transformation is represented by g(x) = f(x) + 2. Common errors include confusing vertical and horizontal shifts: g(x) = f(x - 2) would shift right 2 units, while g(x) = f(x + 2) would shift left 2 units. Remember that transformations outside the function affect y-values (vertical), while transformations inside affect x-values (horizontal).

Question 7

1. $y=1$
2. $x=-1$ (correct answer)
3. $y=-1$
4. $x=1$

Explanation: We need to find the axis of symmetry of a parabola with vertex at (-1, 1) that passes through (0, 3) and (-2, 3). The axis of symmetry of a parabola is a vertical line that passes through the vertex, given by x = h where (h, k) is the vertex. Since the vertex is at (-1, 1), the axis of symmetry is x = -1. The fact that points (0, 3) and (-2, 3) have the same y-value confirms they are equidistant from x = -1. A common error is confusing the axis of symmetry with a horizontal line through the vertex, which would be y = 1.

Question 8

1. $f(x)=-|x|+3$ (correct answer)
2. $f(x)=|x|+3$
3. $f(x)=-|x-3|$
4. $f(x)=|x|-3$

Explanation: We need to find the equation for an upside-down V-shaped graph with peak at (0, 3). The graph forms an inverted V with vertex at (0, 3) and passes through (-4, -1) and (4, -1), indicating a downward-opening absolute value function. The general form is f(x) = -a|x| + 3 where a > 0. Using point (4, -1): -1 = -a|4| + 3 = -4a + 3, which gives -4a = -4, so a = 1. Therefore, f(x) = -|x| + 3. The negative coefficient flips the V upside down, and the +3 shifts it up. A common error is forgetting the negative sign, which would give f(x) = |x| + 3, an upward-opening V. Always verify the coefficient sign matches the orientation shown in the graph.

Question 9

1. $-4$ (correct answer)
2. $-2$
3. $0$
4. $4$

Explanation: We need to find the minimum value of an upward-opening parabola with vertex at (-2, -4). The graph shows a parabola opening upward with x-intercepts at -4 and 0, and vertex at (-2, -4). For an upward-opening parabola, the vertex represents the minimum point, so the minimum value is the y-coordinate of the vertex, which is -4. We can verify this is correct: the vertex x-coordinate (-2) is indeed the midpoint between the x-intercepts: (-4 + 0)/2 = -2. A common error is reporting the x-coordinate of the vertex (-2) as the minimum value, but we need the y-coordinate. For upward-opening parabolas, always identify the vertex's y-coordinate as the minimum value.

Question 10

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

Explanation: We need to find a point on g(x) = f(x - 3) where f(x) passes through (0, 1) and (2, 5). The transformation g(x) = f(x - 3) shifts the graph of f(x) right by 3 units, so every point (a, b) on f becomes (a + 3, b) on g. Since (0, 1) is on f(x), the corresponding point on g(x) is (0 + 3, 1) = (3, 1). We can verify: if f(0) = 1, then g(3) = f(3 - 3) = f(0) = 1. A common error is shifting left instead of right, which would give (-3, 1). Remember that f(x - 3) shifts right by 3, while f(x + 3) shifts left by 3.

Question 11

1. $(-3,3)$
2. $(3,-3)$ (correct answer)
3. $(3,3)$
4. $(2,-1)$

Explanation: We need to find a point on g(x) = -f(x) where f(x) is a parabola with vertex at (2, 1) passing through (3, 3). The transformation g(x) = -f(x) reflects the graph of f(x) across the x-axis, changing the sign of all y-values while keeping x-values the same. Since (3, 3) is on f(x), we have f(3) = 3, so g(3) = -f(3) = -3. Therefore, (3, -3) is on g(x). A common error is reflecting across the y-axis (which would change x-values) instead of the x-axis. Remember that -f(x) negates y-values only, creating a reflection across the x-axis.

Question 12

1. $x\ge -1$ (correct answer)
2. $x\le -1$
3. All real $x$
4. $x\ge 0$

Explanation: We need to find the domain of a square root function with endpoint at (-1, 0). The graph shows a square root curve starting at (-1, 0) and extending to the right, existing only for x ≥ -1. For square root functions, the domain is restricted to where the expression under the radical is non-negative. Since the curve starts at x = -1 and continues to the right, the domain is x ≥ -1. A common error is misreading the endpoint location or thinking the domain starts at x = 0 because that's where the basic square root function starts. When finding domains of radical functions, identify the leftmost point where the function begins.

Question 13

1. $f(x)=\sqrt{x-1}-2$ (correct answer)
2. $f(x)=\sqrt{x+1}-2$
3. $f(x)=\sqrt{x-1}+2$
4. $f(x)=-\sqrt{x-1}-2$

Explanation: We need to find the equation of a square root function starting at (1, -2). The graph shows a square root curve with its starting point (domain restriction) at (1, -2), indicating a horizontal shift right 1 unit and vertical shift down 2 units. The general form is f(x) = a√(x - h) + k where (h, k) is the starting point, giving f(x) = a√(x - 1) - 2. Since the curve passes through (5, 0), we substitute: 0 = a√(5 - 1) - 2, which gives 0 = 2a - 2, so a = 1. Therefore, f(x) = √(x - 1) - 2. A common error is shifting left instead of right, writing √(x + 1) when the starting point is at x = 1. Remember that √(x - 1) shifts the basic square root function right by 1 unit.

Question 14

1. $f(x)=|x+2|+1$ (correct answer)
2. $f(x)=|x-2|+1$
3. $f(x)=|x+2|-1$
4. $f(x)=-|x+2|+1$

Explanation: We need to find the equation of an absolute value function with vertex at (-2, 1). The V-shaped graph has its vertex (sharp point) at (-2, 1) and opens upward, indicating a positive coefficient. The general form is f(x) = a|x - h| + k where (h, k) is the vertex, so we have f(x) = a|x - (-2)| + 1 = a|x + 2| + 1. Since the graph passes through (0, 3), we substitute: 3 = a|0 + 2| + 1, which gives 3 = 2a + 1, so a = 1. Therefore, f(x) = |x + 2| + 1. A common mistake is confusing the shift direction: |x + 2| shifts left 2 units, not right. When the vertex is at x = -2, you need x + 2 inside the absolute value, not x - 2.

Question 15

1. $f(x)=(x-1)^2+4$
2. $f(x)=-(x-1)^2-4$
3. $f(x)=(x-1)^2-4$ (correct answer)
4. $f(x)=(x+1)^2-4$

Explanation: We need to identify the equation of an upward-opening parabola with vertex at (1, -4). The graph shows the parabola opens upward (positive leading coefficient) with vertex at (1, -4), so the vertex form is f(x) = a(x - 1)² - 4 where a > 0. Since the parabola passes through (0, -3), we can substitute: -3 = a(0 - 1)² - 4, which gives -3 = a - 4, so a = 1. Therefore, f(x) = (x - 1)² - 4. Common errors include using a negative coefficient (which would flip the parabola downward) or shifting the vertex incorrectly by writing (x + 1)² instead of (x - 1)². Remember that in vertex form f(x) = a(x - h)² + k, the vertex is at (h, k), not (-h, k).

Question 16

1. $f(x)=\sqrt{x-2}$ (correct answer)
2. $f(x)=\sqrt{x+2}$
3. $f(x)=\sqrt{x-2}+2$
4. $f(x)=2\sqrt{x-2}$

Explanation: We need to identify a square root function starting at (2, 0) and passing through (3, 1) and (6, 2). The general form is f(x) = a√(x - h) + k, where (h, k) is the starting point, giving us f(x) = a√(x - 2) + 0 = a√(x - 2). To verify a = 1, check point (3, 1): 1 = a√(3 - 2) = a√1 = a, so a = 1. Let's verify with (6, 2): f(6) = √(6 - 2) = √4 = 2 ✓. Therefore, f(x) = √(x - 2). Common mistakes include using √(x + 2) (shifting left instead of right) or adding an unnecessary vertical shift.

Question 17

1. No, it fails at $x=4$
2. No, it is vertical
3. Yes, it is a function (correct answer)
4. No, it fails at $x=0$

Explanation: The question determines if the graphed relation is a function of x using the vertical line test. The graph consists of a line segment from (-4,-1) to (0,3) continuing linearly to (4,7), forming a single straight line without branches or loops. Since no vertical line intersects the graph more than once, it passes the vertical line test and is a function. This test connects the visual uniqueness of y for each x to the function definition. A key error might be thinking it fails at x=0 due to the segment endpoint, but continuity doesn't affect the test. Another mistake could be calling it vertical if slopes are miscalculated, but it's slanted. Apply the vertical line test systematically across the domain to confirm function status on graphs.

Question 18

1. $f(x)=-x(x+4)$
2. $f(x)=x(x+4)$ (correct answer)
3. $f(x)=(x-4)x$
4. $f(x)=(x+2)^2-4$

Explanation: We need to find the equation of an upward-opening parabola with x-intercepts at x = -4 and x = 0, and vertex at (-2, -4). Since the parabola has x-intercepts at -4 and 0, we can write f(x) = ax(x + 4). To find a, we use the vertex (-2, -4): -4 = a(-2)(-2 + 4) = a(-2)(2) = -4a, so a = 1. Therefore, f(x) = x(x + 4). The parabola opens upward since a = 1 > 0, which matches the description. A common error is using the wrong sign for the leading coefficient, which would make the parabola open downward instead.

Question 19

1. Shift left 4 units
2. Shift right 4 units (correct answer)
3. Shift up 4 units
4. Reflect over $x$-axis

Explanation: The question asks which transformation turns f(x) = √x into g(x) = √(x-4), based on the graphs. The original graph starts at (0,0) and increases rightward, while the new one begins at (4,0) and follows similarly, showing a horizontal shift. The equation √(x-4) shifts the graph right by 4 units, as the input requires +4 to start the square root at zero. This algebraic change corresponds to the visual delay in the starting point on the x-axis. A common error is confusing it with a left shift, which would use √(x+4) instead. Another mistake might be seeing it as a vertical shift if the new origin is misread. For square root transformations, note how changes inside the root affect horizontal position to identify shifts accurately.

Question 20

1. $y=|x|-1$
2. $y=|x-2|-1$ (correct answer)
3. $y=|x+2|-1$
4. $y=-|x-2|-1$

Explanation: The question seeks the equation for the graphed absolute value function with given vertex and points. The graph is V-shaped with vertex at (2,-1), passing through (0,1) and (4,1), symmetric about x=2 and opening upward. The form y = |x - 2| - 1 shifts right by 2 and down by 1, verified at x=0: |0-2| - 1 = 1, and x=4: |4-2| - 1 = 1. This links the visual vertex shift to the absolute value adjustments. A key error is wrong horizontal shift, like |x+2| for left instead of right. Another mistake could be adding a negative sign, opening it downward and missing the points. For absolute values, use the vertex to set the form and test points to confirm shifts.