Algebra - Exponential functions and horizontal asymptotes-extended

Given the function $f(x) = 3 \cdot 2^x - 5$, determine the horizontal asymptote, the $y$-intercept, and whether the function represents growth or decay.

1. The horizontal asymptote is $y = k$. Here $k = -5$, so the asymptote is $y = -5$.
2. To find the $y$-intercept, set $x = 0$: $f(0) = 3 \cdot 2^0 - 5 = 3(1) - 5 = -2$. The intercept is $(0, -2)$.
3. Since the base $b = 2$ and $2 > 1$, the function represents exponential growth.

The value of $k$ shifted the natural asymptote $y=0$ down to $y=-5$. Since the base is greater than 1, the values of $y$ increase as $x$ increases.

Find the equation of an exponential function of the form $y = a \cdot b^x + k$ that has a horizontal asymptote at $y = 10$, a $y$-intercept at $(0, 15)$, and passes through the point $(1, 20)$.

1. From the horizontal asymptote, we know $k = 10$. The equation is $y = a \cdot b^x + 10$.
2. Use the $y$-intercept $(0, 15)$: $15 = a \cdot b^0 + 10 \Rightarrow 15 = a(1) + 10 \Rightarrow a = 5$.
3. Now the equation is $y = 5 \cdot b^x + 10$. Use the point $(1, 20)$: $20 = 5 \cdot b^1 + 10 \Rightarrow 10 = 5b \Rightarrow b = 2$.
4. The final equation is $y = 5 \cdot 2^x + 10$.

Identify the parameters in order: $k$ (from asymptote), then $a$ (from $y$-intercept), then $b$ (from the second point).

Consider the function $g(x) = -2 \cdot (0.5)^x + 4$. State the range of the function.

1. The horizontal asymptote is $y = 4$.
2. The coefficient $a$ is $-2$. Since $a$ is negative, the graph is reflected across the horizontal line and exists below the asymptote.
3. Therefore, the range is $y < 4$.

For any $f(x) = a \cdot b^x + k$, if $a$ is negative, the outputs will always be less than the vertical shift $k$ because $b^x$ is always positive.

$1. \text{ Identify } k: k = -3. \text{ The horizontal asymptote is } y = -3.$ $2. \text{ Find y-intercept: } f(0) = 2 \cdot 0.5^0 - 3 = 2(1) - 3 = -1. \text{ Intercept is } (0, -1).$ $3. \text{ Determine growth/decay: } b = 0.5, \text{ which is } < 1, \text{ so it is decay approach } y = -3 \text{ as } x \to \infty.$

The graph starts high on the left, crosses the $y$-axis at $-1$, and flattens out towards the line $y = -3$ as $x$ increases.

$1. \text{ From the asymptote, } k = 3. \text{ Thus, } y = a \cdot b^x + 3.$ $2. \text{ Use } (0, 4): 4 = a \cdot b^0 + 3 \implies 4 = a(1) + 3 \implies a = 1.$ $3. \text{ Use } (1, 6): 6 = 1 \cdot b^1 + 3 \implies 6 = b + 3 \implies b = 3.$ $4. \text{ The equation is } y = 1 \cdot 3^x + 3, \text{ or } y = 3^x + 3.$

We first use the asymptote to find $k$, then the $y$-intercept to find $a$, and finally a second point to solve for the base $b$.