Tallying graph and options to eliminate options

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Many graph questions show a curve and give function answer choices, then ask which one models the graph.
You read points off the graph and plug their xx-values into each choice to see which output matches. Here is a complete example, where the graph passes through (0,4)(0, 4) and (1,12)(1, 12).
tally_options.png
Which function is it?
(A) f(x)=4⋅3xf(x) = 4 \cdot 3^x
(B) f(x)=4⋅2xf(x) = 4 \cdot 2^x
Read a point and plug in its xx-value.
At x=0x = 0, both choices give 44, so that point cannot separate them.
At x=1x = 1, the graph shows 1212: choice (A) gives 4⋅3=124 \cdot 3 = 12, matching, while (B) gives 4⋅2=84 \cdot 2 = 8, which misses, so (A) is the model.
Save time by eliminating with obvious features first.
If the graph rises, drop any decaying choice; if the yy-intercept is 44, drop any choice that does not give 44 at x=0x = 0.
Then test the survivors at a point where they disagree, and keep the one that matches.

Worked examples

The graph of g(x)g(x) passes through (0,11)(0, 11) and (2,2)(2, 2).
tally_ex1.png
Which function is it?
(A) g(x)=12(12)x−1g(x) = 12(\tfrac{1}{2})^x - 1
(B) g(x)=15(12)x−4g(x) = 15(\tfrac{1}{2})^x - 4
Both choices give about 1111 at x=0x = 0, so read the second point: at x=2x = 2 the graph shows 22.
Choice (A) gives 12⋅14−1=212 \cdot \tfrac{1}{4} - 1 = 2, matching the graph, while (B) gives 15⋅14−4=−0.2515 \cdot \tfrac{1}{4} - 4 = -0.25, which misses, so (A) is correct.
A scatterplot shows salary rising with years of experience; at x=2x = 2, the salary is about 2525 thousand.
tally_ex2.png
Which function models it?
(A) f(x)=20+3.5xf(x) = 20 + 3.5x
(B) f(x)=20(1.1)xf(x) = 20(1.1)^x
Read the graph at x=2x = 2, where the salary is about 2525.
Choice (A) gives 20+3.5(2)=2720 + 3.5(2) = 27, too high, while (B) gives 20(1.1)2≈2420(1.1)^2 \approx 24, close to the graph, so (B) is correct.
The graph of f(x)f(x) passes through (0,500)(0, 500) and (1,200)(1, 200).
tally_ex3.png
Which function is it?
(A) f(x)=500(0.6)xf(x) = 500(0.6)^x
(B) f(x)=500(0.4)xf(x) = 500(0.4)^x
Both choices give 500500 at x=0x = 0, so read the point at x=1x = 1, where the graph shows 200200.
Choice (A) gives 500(0.6)=300500(0.6) = 300, too high, while (B) gives 500(0.4)=200500(0.4) = 200, matching the graph, so (B) is correct.

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