understanding elimination of options by reading graphs

On a graph-modelling question you are asked which function fits a curve or scatterplot.
Before testing lots of points, use the y-intercept — the value at $x = 0$ — because it is the easiest point to read and to compute.

Substituting $x = 0$ is quick because most terms collapse.
For $f(x) = 20 \times 1.1^x$, $f(0) = 20 \times 1 = 20$, and for $f(x) = 20 + 1.1^x$, $f(0) = 20 + 1 = 21$.
If the graph clearly starts at $20$, the second choice is already out.

The same idea catches choices with the wrong minimum or starting value.
A choice like $f(x) = 20 \times 1.1x$ gives $f(0) = 0$, so its lowest value is $0$, not the $\$20{,}000$ the graph shows.
That mismatch lets you eliminate it without checking any other point.

After the $x = 0$ check, use another easy point only among the survivors.
Because one well-chosen point often removes two or three choices, you rarely test all four in full.
Read the value the graph gives, compute each remaining choice there, and keep the match.

A graph starts at $f(0) = 30$. Which function fits?
(A) $f(x) = 30 \times 1.2^x$
(B) $f(x) = 30 + 1.2^x$
At $x = 0$, (A) gives $30 \times 1 = 30$, matching the graph.
(B) gives $30 + 1 = 31$, which misses, so (A) is correct.

The graph's minimum value is $15$, reached at $x = 0$. Which choice can you eliminate?
(A) $f(x) = 15 \times 1.2x$
(B) $f(x) = 15 + 2x$
At $x = 0$, (A) gives $15 \times 1.2 \times 0 = 0$, so its minimum is $0$, not $15$.
(B) gives $15$, which matches, so eliminate (A).

A curve passes through $(0, 5)$. Which function fits?
(A) $f(x) = 5(2)^x$
(B) $f(x) = 3(2)^x$
At $x = 0$, (A) gives $5 \times 1 = 5$, matching the intercept.
(B) gives $3$, which misses, so (A) is correct.