Graphing Functions - PSAT Math

A coordinate plane shows the graph of $f(x)$ as a V-shaped absolute value function with vertex labeled at $(-2,1)$ . The arms rise with slope 1 to the left and right, and the point $(-1,2)$ lies on the right arm. What is the graph of $f(x)-3$ ?
Answer: Shift down 3 units The question asks for the transformation representing the graph of f(x) - 3, where f(x) is a V-shaped absolute value function with vertex at (-2,1) and arms extending with slope 1 on the right and effectively rising on both sides. The graph shows the vertex and a point (-1,2) on the right arm, illustrating the characteristic V-shape shifted left by 2 and up by 1 from the standard |x|. Subtracting 3 shifts the entire graph down by 3 units, moving the vertex to (-2,-2) while preserving the shape and slopes, linking the algebraic subtraction to a vertical translation. This matches the described transformation. A key error is confusing vertical shifts with horizontal ones, like choice A, or reflections as in choice D, which would invert the V. When identifying transformations from graphs, focus on how vertex coordinates change to determine shifts or reflections.
The coordinate plane shows the graph of $f(x)$ as a V-shaped absolute value function with vertex at $(-2,1)$ . The graph passes through $(0,5)$ and $(-4,5)$ . Which graph would represent $g(x)=f(x)-3$ ? Use the shift shown by subtracting 3 from the function values.
Answer: Shift down 3 units This question requires identifying the transformation that results in g(x) = f(x) - 3 for the given absolute value function f(x) with vertex at (-2, 1) and passing through (0, 5) and (-4, 5). The graph shows a V-shaped function opening upward, symmetric about x = -2, with the vertex at (-2, 1) and arms extending to the points provided, indicating a standard absolute value shifted left and up. Subtracting 3 from f(x) lowers every y-value by 3 units, which visually shifts the entire graph downward by 3 units while preserving the shape and x-coordinates of key points. Algebraically, this transformation moves the vertex to (-2, -2) and adjusts points like (0, 5) to (0, 2), connecting the functional change to the vertical translation observed. A key error is mistaking it for a horizontal shift, such as left or right, which would alter x-values instead of y-values. Another common mistake is confusing it with a reflection, which would invert the graph rather than translate it. To strategize, always recall that adding or subtracting a constant outside the function affects vertical position, and test by applying the change to a known point on the graph.
The coordinate plane shows the graph of a parabola opening upward. Its vertex is labeled at $(1,-4)$ , and the graph passes through the point $(0,-3)$ (one unit left of the vertex and one unit higher). Which equation could represent the graphed function?
Answer: $y=(x-1)^2-4$ The question requires identifying the equation of an upward-opening parabola with vertex at (1,-4) and passing through (0,-3). The graph depicts a parabola symmetric about x=1, bottoming at y=-4, and rising to (0,-3) one unit left and one unit up from the vertex. Using vertex form y = $a(x - h)^2$ + k with h=1, k=-4, and substituting (0,-3) gives a=1, resulting in y = $(x-1)^2$ - 4, which connects the vertex shift and positive coefficient to the upward opening. This equation aligns with the graphed points and direction. Errors often arise from incorrect vertex placement, like choice C shifting to (-1,-4), or negative coefficients causing downward opening as in choice B. For parabola graphs, plug in given points to verify the vertex form matches the visual minimum or maximum.
A coordinate plane graph shows the function $f(x)$ as a line passing through $(0,2)$ and $(4,0)$ . The function $g(x)$ is defined by $g(x)=f(x)-3$ . Which graph best represents $g(x)$ compared to $f(x)$ ?
Answer: Shift down 3 units We need to determine how g(x) = f(x) - 3 transforms the graph of f(x). The transformation g(x) = f(x) - 3 subtracts 3 from every y-value of f(x), which shifts the entire graph down by 3 units. Since f(x) passes through (0, 2), g(x) will pass through (0, 2 - 3) = (0, -1). Similarly, since f(x) passes through (4, 0), g(x) will pass through (4, 0 - 3) = (4, -3). This confirms a downward shift of 3 units. A common error is thinking that subtracting from the function shifts it up, but remember: adding to f(x) shifts up, subtracting shifts down. For quick verification on tests, pick any point on the original graph and apply the transformation.
A coordinate plane shows the graph of $f(x)=\sqrt{x}$ starting at $(0,0)$ and passing through $(4,2)$ . A new graph $g(x)$ is shown on the same axes, starting at $(1,0)$ and passing through $(5,2)$ . Which equation represents $g(x)$ ?
Answer: $g(x)=\sqrt{x-1}$ The question asks for the equation of g(x), a square root graph starting at (1,0) and passing through (5,2), compared to f(x)=√x through (0,0) and (4,2). The graph of g(x) appears as f(x) shifted right by 1 unit, with the starting point moved from (0,0) to (1,0) and the curve following similarly. Replacing x with x-1 in √x gives g(x)=√(x-1), which starts at x=1 and matches (5,√4=2), connecting the horizontal shift to the visual translation. This represents the graphed function. Common errors include vertical shifts like choices A or D, which wouldn't align the starting points correctly. When comparing transformed graphs, identify changes in domain or key points to determine shift type.
The coordinate plane shows a line that crosses the $y$ -axis at $-1$ and passes through the point $(4,3)$ . Which equation matches the graphed line? Several choices have a similar slope, so use both the intercept and the plotted point.
Answer: $y=x-1$ This question requires the equation of a line crossing the y-axis at -1 and passing through (4, 3). The graph depicts a straight line with y-intercept at (0, -1) and another point at (4, 3), allowing calculation of the slope. The slope m = (3 - (-1))/(4 - 0) = 4/4 = 1, leading to y = x - 1 in slope-intercept form, which matches both points algebraically and visually. This connects the intercept and slope to the line's position and steepness on the graph. A key error is miscalculating the slope, such as to 3/4 or 1/2, resulting in lines that miss (4, 3). Another mistake is using y = x + 1, which has the wrong intercept. Always use both the intercept and a given point to verify the slope when multiple options have similar forms.
A coordinate plane graph shows a function with a horizontal asymptote at $y=2$ . The curve passes through $(0,3)$ and decreases toward $y=2$ as $x$ increases. Which equation best matches the graph? (If you miss the asymptote, you might choose an unshifted exponential.)
Answer: $y=\left(\tfrac12\right)^x+2$ The question requires the equation for a graph with horizontal asymptote y=2, passing through (0,3), decreasing toward 2 as x increases. The curve approaches y=2 from above without crossing, starting higher at x=0 and flattening rightward, indicating exponential decay shifted up. y = $(1/2)^x$ + 2 fits: at x=0, 1 + 2 = 3; as x→∞, $(1/2)^x$ →0, y→2, matching decay and asymptote. The base less than 1 ensures decreasing, connecting visual approach to algebraic form. Missing the asymptote might lead to unshifted A or D; sign errors select C, which goes below. Confusing growth with decay picks increasing options. Check y-intercept and long-term behavior to confirm exponentials.
The graph of the exponential function $f(x) = a \cdot b^x + c$ is shown. What is the value of $a + b + c$ ?
Answer: $\tfrac{1}{2}$ The horizontal asymptote is $y = -2$ , so $c = -2$ . The y-intercept is $f(0) = a + c = 0$ , giving $a = 2$ . Another point: $f(1) = 2b - 2 = -1$ , so $b = 1/2$ . Then $a+b+c = 2 + 1/2 - 2 = 1/2$ . A reverses signs; C forgets to subtract c; D misreads the asymptote.
The coordinate plane shows a line segment graphed for a function $f$ . The segment starts at a closed dot at $(-4,3)$ and ends at an open dot at $(2,-3)$ . What is the domain of the graphed function? Read the endpoint symbols carefully to avoid including an excluded value.
Answer: $[-4,2)$ This question asks for the domain of a function represented by a line segment from a closed dot at (-4, 3) to an open dot at (2, -3). The graph shows a straight line connecting these points, with the closed dot indicating inclusion of x = -4 and the open dot excluding x = 2. The domain, or set of x-values, thus ranges from -4 to 2, inclusive at -4 and exclusive at 2, expressed in interval notation as [-4, 2). This connects the visual endpoint symbols to the algebraic inclusion or exclusion in the domain. A key error is misreading the dots, such as including 2 for a closed interval like [-4, 2] or excluding -4 for (-4, 2). Another mistake involves confusing domain with range, focusing on y-values instead. Always check endpoint symbols carefully and use test points near the ends to confirm inclusion when determining domains from graphs.
Based on the graph of $y = f(x)$ shown, how many solutions are there to the equation $f(f(x)) = 0$ on the interval shown?
Answer: Five First, find values $u$ where $f(u) = 0$ : from the graph, $u = -3$ and $u = 2$ . Then solve $f(x) = -3$ and $f(x) = 2$ . From the graph, $f(x) = -3$ has 2 solutions and $f(x) = 2$ has 3 solutions. Total: 5 solutions. A counts only direct zeros; B counts $f(x)=2$ only; C misses one intersection.
A coordinate plane graph shows a straight line passing through the labeled points $(-2,3)$ and $(4,-1)$ . The line crosses the $y$ -axis between 1 and 2, and it slopes downward from left to right. Which equation matches the graphed line?
Answer: $y=-\tfrac{2}{3}x+\tfrac{5}{3}$ The question asks for the equation of a straight line passing through points (-2,3) and (4,-1), with the line crossing the y-axis between 1 and 2 and sloping downward. The graph shows a line with a negative slope, intersecting the y-axis at approximately 1.67, and connecting the given points. To find the equation, calculate the slope as m = (-1 - 3)/(4 - (-2)) = -4/6 = -2/3, confirming the downward trend, then use point-slope form with (-2,3) to find the y-intercept b = 5/3, yielding y = -(2/3)x + 5/3. This matches the visual y-intercept between 1 and 2 and the downward slope from left to right. A common error is miscalculating the slope sign, leading to positive slope options like choice A, or incorrect intercept signs like choice D. When graphs provide points, always verify the slope and intercept algebraically to match the visual features.
Based on the graph shown, which statement best compares $f$ and $g$ ?
Answer: $f$ and $g$ have equal maximum values, but $f$ has a lower minimum. Graph shows both curves peak at y=4, but f dips to –2 while g only dips to 0. Hence equal maxima, f lower minimum. Choice A reverses, B wrong on maxima, D incorrect.
The graphs of $y = f(x)$ and $y = g(x)$ are shown on the same coordinate plane. For how many integer values of $x$ in $-4 \le x \le 4$ is $f(x) > g(x)$ ?
Answer: Four To find where f(x) > g(x), we need to identify the x-values where the graph of f lies above the graph of g. Checking each integer from -4 to 4: f(x) > g(x) occurs at x = -4, -3, 2, and 3. At the other integer values, either g(x) > f(x) or the functions are equal. Therefore, there are 4 integer values where f(x) > g(x).
The graph of the rational function $f(x) = \dfrac{ax + b}{x + c}$ is shown. What is the value of $a + b + c$ ?
Answer: $3$ The vertical asymptote is $x = -2$ , so $c = 2$ . The horizontal asymptote is $y = 2$ , so $a = 2$ . The x-intercept is at $x = \tfrac{1}{2}$ : $0 = \tfrac{2(1/2) + b}{1/2 + 2}$ , so $1 + b = 0$ , $b = -1$ . Then $a+b+c = 2 - 1 + 2 = 3$ . A forgets to include $a$ ; C mis-signs $c$ ; D mis-signs $b$ .
A coordinate plane shows the graph of $f(x)$ as an increasing line that crosses the $y$ -axis at $-1$ and passes through $(2,3)$ . If $g(x)=f(x-3)$ , which change occurs to the graph when moving from $f$ to $g$ ?
Answer: Shift right 3 units The question asks how g(x) = f(x - 3) transforms the graph of f(x), where f(x) is a line crossing the y-axis at -1 and passing through (2, 3). When we have g(x) = f(x - 3), this represents a horizontal shift of the graph. The transformation x - 3 inside the function shifts the graph right by 3 units (counterintuitively, subtracting inside shifts right). Every point (x, y) on f becomes (x + 3, y) on g. For example, the y-intercept (0, -1) becomes (3, -1), and the point (2, 3) becomes (5, 3). A common error is thinking that f(x - 3) shifts left because of the minus sign, but remember: inside the function, minus means right.
A coordinate plane shows a parabola opening upward with its lowest point at $(1,-4)$ . The curve crosses the $x$ -axis at $x=-1$ and $x=3$ , and it crosses the $y$ -axis at $y=-3$ . Which equation matches the graphed function? Choose the equation whose graph would have the same vertex and intercepts as shown.
Answer: $y=(x+1)(x-3)$ This question asks for the equation of a parabola that opens upward with a vertex at (1, -4), x-intercepts at x = -1 and x = 3, and a y-intercept at y = -3. The graph depicts a U-shaped curve with its minimum point at (1, -4), crossing the x-axis at (-1, 0) and (3, 0), and intersecting the y-axis at (0, -3). To find the matching equation, recognize that the x-intercepts indicate roots at x = -1 and x = 3, so the factored form is y = a(x + 1)(x - 3); substituting the vertex or y-intercept confirms a = 1, yielding y = (x + 1)(x - 3). Expanding this gives y = x² - 2x - 3, which matches the vertex form y = (x - 1)² - 4 after completing the square, confirming the algebraic representation aligns with the visual features. A common error is selecting y = (x - 1)² - 3, which has a vertex at (1, -3) instead of (1, -4), or choosing the negative version that opens downward. Another mistake is confusing the vertex form's constant term, leading to incorrect minimum value. When multiple forms represent the same graph, verify by plugging in key points like intercepts to ensure consistency.
A coordinate plane graph shows the function $f(x)$ as a downward-opening parabola. The vertex is labeled at $(2,3)$ , and the parabola passes through $(1,2)$ and $(3,2)$ . Which equation matches the graphed function $f(x)$ ?
Answer: $f(x)=-(x-2)^2+3$ We need to find the equation of a downward-opening parabola with vertex at (2, 3). The vertex form of a parabola is f(x) = a(x - h)² + k, where (h, k) is the vertex and a determines the direction and width. Since the parabola opens downward, a must be negative. With vertex (2, 3), we have f(x) = a(x - 2)² + 3. To find a, we use the point (1, 2): substituting gives 2 = a(1 - 2)² + 3, which simplifies to 2 = a(1) + 3, so a = -1. Therefore, f(x) = -(x - 2)² + 3. The key insight is recognizing that a downward-opening parabola requires a negative coefficient, eliminating choices B and any with positive leading terms.
The graph of $y = g(x)$ is shown in the xy-plane. If $h(x) = g(2x) - 3$ , which of the following is the y-intercept of the graph of $y = h(x)$ ?
Answer: $-1$ The y-intercept of $h(x)$ occurs at $x = 0$ : $h(0) = g(0) - 3$ . From the graph, $g(0) = 2$ , so $h(0) = 2 - 3 = -1$ . Distractor A assumes $g(0) = -2$ . Distractor B ignores $g(0)$ . Distractor D forgets to subtract 3.
A coordinate plane shows the graph of $f(x)=\sqrt{x}$ starting at $(0,0)$ and passing through $(4,2)$ . Which graph represents $g(x)=\sqrt{x-1}$ ? Interpret the transformation on the coordinate plane; shifting the wrong direction would still look similar near the origin.
Answer: Shift right 1 unit The question seeks the transformation representing $g(x) = \sqrt{x - 1}$ from the graph of $f(x) = \sqrt{x}$ starting at $(0, 0)$ and passing through $(4, 2)$ . The original graph is a half-parabola opening rightward from the origin, with the point $(4, 2)$ indicating the standard square root shape. Replacing x with x - 1 in the function shifts the graph right by 1 unit, moving the starting point to $(1, 0)$ and adjusting points accordingly, connecting the algebraic substitution to the horizontal translation visible. This preserves the shape but relocates it along the x-axis. A common error is confusing it with a left shift, which would use x + 1, or a vertical shift, which affects y-values instead. Reflecting across the y-axis would use $\sqrt{-x}$ , flipping the graph leftward. To avoid mistakes, remember that subtracting inside the function shifts right, and verify by checking a transformed point like the origin moving to $(1, 0)$ .
The coordinate plane shows a decreasing exponential function with horizontal asymptote $y=0$ . The curve passes through $(0,8)$ and $(1,4)$ . What is the value of the function at $x=2$ ? Use the plotted points to determine the pattern rather than guessing from the curve's shape.
Answer: $2$ This question requires finding the function value at x = 2 for a decreasing exponential with asymptote y = 0, passing through (0, 8) and (1, 4). The graph shows a curve approaching y = 0 from above as x increases, starting higher on the left and decreasing through the given points, indicative of exponential decay. Using y = a * $b^x$ , (0, 8) gives a = 8, and (1, 4) gives 8b = 4 so b = 1/2; at x = 2, y = 8 * $(1/2)^2$ = 2, connecting the decay factor to the pattern in points. This algebraic model matches the visual halving each unit increase in x. A key error is assuming a base like 2 or 4 without the decay, leading to growth instead of decrease. Guessing from shape without calculation might suggest 1, 4, or 16, missing the pattern. Use consecutive points to find the common ratio and extend it to the desired x-value for accuracy.

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