Algebra - Exponential Functions and Horizontal Asymptotes

Identify the horizontal asymptote and the $y$-intercept for the function $f(x) = 3(2^x) + 5$.

Horizontal Asymptote: $y = 5$; $y$-intercept: $(0, 8)$.

Comparing $f(x) = 3(2^x) + 5$ to the general form $f(x) = a \cdot b^x + k$, we see that $k = 5$. Therefore, the horizontal asymptote is $y = 5$. To find the $y$-intercept, we substitute $x = 0$: $f(0) = 3(2^0) + 5 = 3(1) + 5 = 8$. Thus, the intercept is at the point $(0, 8)$.

Determine if the function $g(x) = 10(0.75)^x - 2$ represents growth or decay, and state its horizontal asymptote.

The function represents exponential decay; Horizontal Asymptote: $y = -2$.

Since the base $b = 0.75$ is between $0$ and $1$ ($0 < 0.75 < 1$), the function represents exponential decay. The constant term $k$ is $-2$, which means the graph approaches the line $y = -2$ as $x$ increases.

Given the function $h(x) = -2(3^x) + 4$, find the value of $h(x)$ when $x = 2$ and state the equation of the horizontal asymptote.

$h(2) = -14$; Horizontal Asymptote: $y = 4$.

To find $h(2)$, substitute $x = 2$ into the equation: $h(2) = -2(3^2) + 4 = -2(9) + 4 = -18 + 4 = -14$. The horizontal asymptote is determined by the constant term $k = 4$, so the equation is $y = 4$.

1. Identify $k$: Here, $k = -3$. The horizontal asymptote is $y = -3$.
2. Find the $y$-intercept: Set $x = 0$. $f(0) = 2^0 - 3 = 1 - 3 = -2$. The $y$-intercept is $(0, -2)$.
3. Behavior: Since $b = 2$ and $2 > 1$, the function is increasing (growth).

The graph shifts 3 units down from the parent function $y=2^x$, moving the asymptote from $y=0$ to $y=-3$.

1. Range: Since $a = 4$ (positive) and $k = 1$, the function stays above the asymptote. Range: $y > 1$.
2. Calculate $g(2)$: $g(2) = 4(0.5)^2 + 1$ $g(2) = 4(0.25) + 1$ $g(2) = 1 + 1 = 2$.

Because the base is $0.5$, the function decays toward the horizontal line $y=1$. At $x=2$, the value is exactly $2$.