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.

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.

The question seeks the equation of an exponential function passing through (0, 2) and (1, 4), approaching y = 0 as x decreases. The graph illustrates a curve starting near the x-axis on the left, rising through (0, 2) and (1, 4), characteristic of exponential growth with a horizontal asymptote at y = 0. To match this, use the general form y = a * $b^x$; the point (0, 2) gives a = 2, and (1, 4) gives 2 * b = 4, so b = 2, resulting in y = 2 * $2^x$, which aligns with the visual growth rate. This connects the initial value and doubling factor to the plotted points algebraically. A common error is selecting y = $2^x$, which starts at (0, 1) instead of (0, 2), or y = $4^x$, which also starts at 1 but grows faster. Mistaking the base for 4 while keeping the multiplier can lead to y = 2 * $4^x$, which passes through (0, 2) but reaches (1, 8), not (1, 4). For such questions, plug in the given points to verify the equation rather than relying solely on the curve's shape.

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.

$$f^{-1}(2)$$ is the x-value where $$f(x) = 2$$. From the graph, $$f(x) = 2$$ at $$x = 4$$. $$f^{-1}(-1)$$ is where $$f(x) = -1$$, which occurs at $$x = -1$$. Sum: $$4 + (-1) = 3$$. A confuses inverse with negation; B reads coordinates as (input, input); D swaps inputs and outputs incorrectly.

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.

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.

The graph shows zeros at x = -2 (crosses axis), x = 1 (touches axis, indicating multiplicity 2), and x = 4 (crosses axis). This gives the factored form (x+2)(x-1)²(x-4). The end behavior shows that as x → +∞, y → -∞, indicating a negative leading coefficient. Therefore, the equation is p(x) = -¼(x+2)(x-1)²(x-4).

$$f(x) = f(-x)$$ when a point and its reflection across the y-axis have the same y-value. From the graph, this happens at $$x = 0$$ (trivially), and at the two x-values where the graph intersects its reflection. Checking the graph: the values $$x=-3$$ and $$x=3$$ give matching y-values (both 1), and $$x=0$$ works. That gives 3 values. A common error is counting only the nonzero matches (2) or missing the symmetric pair.

$$h(4) = 2f(4/2) = 2f(2)$$. From the graph, $$f(2) = 3$$, so $$h(4) = 2(3) = 6$$. A negates incorrectly; B forgets to multiply by 2; D uses $$f(4)$$ instead of $$f(2)$$.