Graphing Functions - SAT Math

The coordinate plane shows the graph of $f(x)$ , a parabola opening upward with vertex at $(0,0)$ and passing through $(1,2)$ . Which transformation produces the graph of $g(x)=f(x)+3$ ?
Answer: Shift up 3 The question asks about the transformation from f(x) to g(x) = f(x) + 3, where f(x) is a parabola with vertex at (0, 0) passing through (1, 2). The transformation g(x) = f(x) + 3 adds 3 to every y-value of f(x), which shifts the entire graph up by 3 units. This moves the vertex from (0, 0) to (0, 3) and the point (1, 2) to (1, 5). A common error is confusing f(x) + 3 with f(x + 3), where f(x + 3) would shift the graph left by 3 units instead. Remember: adding outside the function shifts vertically, adding inside the function shifts horizontally in the opposite direction.
A line in the coordinate plane has $y$ -intercept $-3$ and $x$ -intercept $2$ . What is the slope of the line?
Answer: $\frac{3}{2}$ The question asks for the slope of the line given its intercepts on the graph. The line crosses the y-axis at (0,-3) and x-axis at (2,0), showing an upward trend from left to right. Calculate slope as m \= (0 - (-3)) / (2 - 0) \= 3/2, using the intercept points. This rise over run connects the visual positive direction to the algebraic slope formula. A common error is swapping points, yielding -3/2 and a negative slope mismatch. Another mistake could be using 2/3 if inverting the fraction. When intercepts are given, plot them mentally and compute slope to avoid sign errors.
On the coordinate plane, the graph of $f(x)$ is a square-root curve that begins at $(1,-2)$ and increases to the right, passing through $(5,0)$ . Which equation matches the graph?
Answer: $f(x)=\sqrt{x-1}-2$ The square root function begins at (1, -2) and passes through (5, 0), indicating a transformation of the parent function f(x) = √x. The starting point at (1, -2) suggests a horizontal shift right 1 unit and vertical shift down 2 units, giving f(x) = √(x - 1) - 2. To verify: at x = 1, f(1) = √(1 - 1) - 2 = 0 - 2 = -2 ✓; at x = 5, f(5) = √(5 - 1) - 2 = √4 - 2 = 2 - 2 = 0 ✓. A common error is using √(x + 1) which would shift left instead of right, or forgetting that the square root function's domain starts where the expression under the radical equals zero. Remember that √(x - h) shifts the starting point to x = h.
Let $f(x)=2x+3$ and $g(x)=kx+1$ . If $(g\circ f)(-4)=5$ , what is $k$ ?
Answer: $-\frac{4}{5}$ Since $f(-4)=-5$ , we have $g(f(-4))=-5k+1=5$ , so $-5k=4$ and $k=-\tfrac{4}{5}$ . The other options come from sign mistakes or inverting the fraction.
Let $f(x) = ax + b$ and $g(x) = x^2$ . If $(f \circ g)(2) = 11$ and $(g \circ f)(2) = 49$ with $f(2) > 0$ , what is the value of $a$ ?
Answer: 2 From $f(g(2))=f(4)=11$ , we have $4a+b=11$ . Also $(g\circ f)(2)=(f(2))^2=49$ and $f(2)>0$ imply $2a+b=7$ , so subtracting gives $a=2$ ; other options follow from taking the negative root or arithmetic slips.
A coordinate plane graph shows the parabola $f(x)=x^2$ (standard shape) and another parabola $g(x)$ that has vertex at $(0,0)$ but is narrower, passing through $(1,4)$ . Which equation could be $g(x)$ ?
Answer: $g(x)=4x^2$ We need to identify which parabola is narrower than f(x) = x² but shares the same vertex at (0, 0). The graph shows g(x) passing through (1, 4) while f(x) = x² passes through (1, 1), indicating g(x) rises more steeply. For a parabola g(x) = ax² with vertex at origin, using point (1, 4): 4 = a(1)² = a, so a = 4. Therefore, g(x) = 4x². A vertical stretch by factor 4 makes the parabola narrower (steeper). Common errors include confusing vertical stretch with horizontal compression or thinking g(x) = x² + 4 (which shifts up) or g(x) = (x + 4)² (which shifts left). For parabolas with the same vertex, compare their widths by looking at y-values for the same x-values.
A coordinate-plane graph shows a line segment from $(-4,-1)$ to $(2,5)$ with solid endpoints. Which statement gives the domain of the graphed function?
Answer: $[-4,2]$ This question asks for the domain of a line segment shown on the graph. The line segment connects (-4, -1) to (2, 5) with solid endpoints, indicating both endpoints are included. The domain consists of all possible x-values, which range from x = -4 to x = 2, inclusive. Since the endpoints are solid (not open circles), we use square brackets to indicate inclusion: [-4, 2]. A common error is using the y-values instead of x-values, which would give [-1, 5], or using parentheses instead of brackets for solid endpoints. Remember that domain refers to the horizontal extent (x-values) of the graph.
A coordinate-plane graph shows an absolute value function $f(x)$ with vertex at $(0,0)$ and passing through $(2,6)$ . Which equation matches the graph?
Answer: $f(x)=3|x|$ The absolute value function has vertex at (0, 0) and passes through (2, 6). The general form is $f(x) = a|x|$ , where a determines the vertical stretch. To find a, substitute the point (2, 6): $6 = a|2| = 2a$ , so $a = 3$ . Therefore, $f(x) = 3|x|$ . This represents a vertical stretch by factor 3 of the parent function $|x|$ . A common error is thinking the point (2, 6) means we add 6 to get $f(x) = |x| + 6$ , but this would give $f(2) = 2 + 6 = 8$ , not 6. Another error is using $a = 1/2$ , giving $f(x) = (1/2)|x|$ , which would make $f(2) = 1$ . Always verify your equation with the given points.
On the coordinate plane, the graph of a circle is shown with center at $(0,0)$ and radius $3$ . Does this graph represent a function of $x$ ?
Answer: No, it fails the vertical line test The question assesses if the graphed circle represents a function of x. The graph is a circle centered at (0,0) with radius 3, symmetric about both axes, intersecting vertical lines within |x|<3 at two points. It fails the vertical line test because for each x in (-3,3), there are two corresponding y-values, one positive and one negative. This visual duplication links to the algebraic impossibility of a single output per input for a function. A common error is claiming it's a function due to symmetry, missing the multi-valued nature. Another mistake might be limiting to x≥0, but the full circle is graphed. Always use the vertical line test on closed shapes like circles to detect non-functions.
Given $g(x)=2x-3$ and $(f \circ g)(x)=x^2-1$ , what is $f(x)$ ?
Answer: $\frac{x^2 + 6x + 5}{4}$ We need $f(2x-3)=x^2-1$ ; let $u=2x-3\Rightarrow x=\frac{u+3}{2}$ , so $f(u)=\left(\frac{u+3}{2}\right)^2-1=\frac{u^2+6u+5}{4}$ and $f(x)=\frac{x^2+6x+5}{4}$ . The other choices ignore the inner shift, divide only by 2, or mishandle the expansion.
The coordinate-plane graph shows a reciprocal function with vertical asymptote at $x=1$ and horizontal asymptote at $y=-2$ . A marked point on the curve is $(2,-1)$ . Which equation matches the graph?
Answer: $y=\frac{1}{x-1}-2$ This question asks us to identify a reciprocal function from its asymptotes and a point. The graph has a vertical asymptote at x = 1 and horizontal asymptote at y = -2, suggesting the form y = 1/(x - 1) + k. Since the horizontal asymptote is at y = -2, we have k = -2, giving us y = 1/(x - 1) - 2. We can verify using the point (2, -1): y = 1/(2 - 1) - 2 = 1/1 - 2 = -1 ✓. Common errors include using the wrong sign in the numerator (which would flip the curve) or confusing the vertical shift direction. For rational functions, the horizontal asymptote equals the constant term when the degrees of numerator and denominator differ by one.
The coordinate-plane graph shows a square-root curve starting at the solid point $(2,-1)$ and increasing to the right, passing through $(6,1)$ . Which equation matches the graph?
Answer: $y=\sqrt{x-2}-1$ This question requires identifying the equation of a square root function from its graph. The curve starts at the solid point (2, -1) and increases to the right, passing through (6, 1). For a square root function y = √(x - h) + k, the starting point is at (h, k), so with starting point (2, -1), we have h = 2 and k = -1, giving us y = √(x - 2) - 1. We can verify with point (6, 1): y = √(6 - 2) - 1 = √4 - 1 = 2 - 1 = 1 ✓. A common error is confusing the horizontal shift direction, using √(x + 2) instead of √(x - 2), which would start the graph at x = -2. Remember that √(x - h) shifts the graph right by h units.
A coordinate plane shows the graph of an exponential function passing through $(0,1)$ , $(1,2)$ , and $(2,4)$ . Which of the following equations matches the graph?
Answer: $y=2^x$ The question seeks the equation matching the graphed exponential function passing through given points. The graph shows a curve starting at $(0,1)$ , rising through $(1,2)$ and $(2,4)$ , indicating exponential growth with base greater than 1. To match the equation, test $y = 2^x$ : at $x=0, 2^0=1$ ; $x=1, 2^1=2$ ; $x=2, 2^2=4$ , fitting perfectly. This connects the visual increasing pattern and y-intercept to the exponential base and form. A key error is selecting a decreasing base like $1/2$ if the growth direction is misread. Another mistake could be choosing a linear option like $y=2x$ , which fits two points but not the third. When identifying exponentials, verify multiple points and confirm if the curve is growing or decaying to select the correct base.
The coordinate-plane graph shows a piecewise function: a line segment from $(-4,2)$ to $(0,0)$ and another line segment from $(0,0)$ to $(4,2)$ , forming a V-shape with vertex at the origin. Which equation matches the graph? Consider whether it is $|x|$ or $\tfrac12|x|$ .
Answer: $y=\tfrac12|x|$ This question asks us to identify the equation of a V-shaped piecewise function. The graph shows line segments from (-4, 2) to (0, 0) and from (0, 0) to (4, 2), forming a V with vertex at the origin. The slope from (-4, 2) to (0, 0) is -2/4 = -1/2, and from (0, 0) to (4, 2) is 2/4 = 1/2. This matches y = (1/2)|x|, which has slopes of ±1/2. We can verify: when x = 4, y = (1/2)|4| = 2 ✓, and when x = -4, y = (1/2)|-4| = 2 ✓. A common error is thinking this is y = |x|, which would have slopes of ±1 and pass through (4, 4) instead of (4, 2).
A coordinate plane graph shows a rational function with a vertical asymptote at $x=2$ and a horizontal asymptote at $y=1$ . The curve passes through the labeled point $(3,2)$ . Which equation matches the graph?
Answer: $y=1+\tfrac{1}{x-2}$ We need to identify the rational function with vertical asymptote at x = 2, horizontal asymptote at y = 1, passing through (3,2). A vertical asymptote at x = 2 means the denominator has a factor (x-2). A horizontal asymptote at y = 1 suggests the function approaches 1 as x approaches infinity, indicating a form like y = 1 + k/(x-2). Using the point (3,2): 2 = 1 + k/(3-2) = 1 + k/1, so k = 1. Therefore, y = 1 + 1/(x-2), which matches option A. Common errors include using (x+2) in the denominator (which would put the asymptote at x = -2) or forgetting to account for the horizontal asymptote. When working with rational functions, identify asymptotes first to determine the basic structure.
A coordinate plane graph shows $f(x)$ as an exponential decay curve passing through $(0,6)$ and $(1,3)$ . The curve approaches $y=0$ as $x$ increases. Which equation matches the graph?
Answer: $f(x)=6\left(\tfrac12\right)^x$ We need to find the equation of an exponential decay function passing through (0, 6) and (1, 3). The graph shows exponential decay (decreasing as x increases) with y-intercept 6, so f(x) = a·bˣ where a = 6. Using point (1, 3): 3 = 6·b¹, which gives b = 1/2. Therefore, f(x) = 6·(1/2)ˣ. The key observation is that from x = 0 to x = 1, the y-value halves from 6 to 3, indicating a base of 1/2. A common error is writing this as exponential growth f(x) = 6·2ˣ, which would increase rather than decrease. For exponential decay, the base must be between 0 and 1.
If $f(x)=x^2+1$ and $g(x)=x+3$ , what is $(g\circ f)(-2)$ ?
Answer: 8 First $f(-2)=(-2)^2+1=5$ , then $g(5)=8$ . Reversing the composition gives $f(g(-2))=2$ , and $1$ comes from using $g(-2)$ alone.
A coordinate plane graph shows $f(x)$ as a square-root curve starting at $(-1,2)$ and increasing to the right, passing through $(0,3)$ and $(3,4)$ . Which equation matches the graph?
Answer: $f(x)=\sqrt{x+1}+2$ The question asks us to identify a square root function starting at (-1, 2) and passing through (0, 3) and (3, 4). The general form is f(x) = a√(x - h) + k, where (h, k) is the starting point, giving us f(x) = a√(x - (-1)) + 2 = a√(x + 1) + 2. To find a, we use point (0, 3): 3 = a√(0 + 1) + 2 = a(1) + 2, so a = 1. Therefore, f(x) = √(x + 1) + 2. Common errors include shifting right instead of left (using √(x - 1)) or forgetting the vertical shift entirely. Remember that the square root function starts where the expression under the radical equals zero.
A coordinate plane shows a rational function with vertical asymptote at $x=-2$ and horizontal asymptote at $y=0$ . The graph passes through $(-1,1)$ . Which equation matches the graph?
Answer: $y=\frac{1}{x+2}$ The question asks for the equation of the graphed rational function with asymptotes and a point. The curve has vertical asymptote at x=-2, horizontal at y=0, and passes through (-1,1), approaching from positive side near the asymptote. The form y = 1/(x+2) matches asymptotes and at x=-1: 1/(-1+2)=1. This algebraic structure reflects the visual asymptote positions and curve direction. A key error is sign in denominator, like x-2 shifting to x=2. Another mistake could be negative numerator, yielding y=-1 at the point. Align asymptotes first in rationals, then confirm direction and points for the correct form.
In the coordinate plane, $f(x)$ is a parabola opening upward with vertex at $(2,1)$ and passing through $(3,3)$ . Let $g(x)=-f(x)$ . Which point must lie on the graph of $g(x)$ ?
Answer: $(3,-3)$ 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). Choice (2, 1) is the vertex of f(x) itself — a student who located the vertex correctly but forgot to negate its y-value. 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.

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