In Algebra 1 you met Exponential Functions: $y = ab^x$ multiplies by $b$ each time $x$ goes up by 1. Now you will move and stretch those graphs with the transformations you already know.
Idea clave
$y = a\,b^{x - h} + k$ is $b^x$ stretched by $a$, shifted right $h$ and up $k$. Its horizontal asymptote is $y = k$. The domain is all real numbers; the range is $(k, \infty)$ if $a > 0$ and $(-\infty, k)$ if $a < 0$. If $b > 1$ the graph shows growth; if $0 < b < 1$, decay.
Por qué funciona
$b^x$ is always positive, and at one end it shrinks toward 0 without ever reaching it: $2^{-10} \approx 0.001$, $2^{-20} \approx 0.000001$. That is the asymptote $y = 0$. Stretching by $a$ keeps values near 0 near 0, and shifting up by $k$ moves the asymptote to $y = k$. A horizontal shift slides the graph sideways but cannot move a horizontal line.
Worked examples
Ejemplo resuelto: Reading a transformed exponential
Describe $y = 3 \cdot 2^{x + 1} - 4$.
- Parent $2^x$; stretch by 3; shift left 1; shift down 4.
- Asymptote: $y = -4$. Since $3 > 0$, the graph is above it: range $(-4, \infty)$.
- $y$-intercept: $3 \cdot 2^{1} - 4 = 2$.
Respuesta: Growth, asymptote $y = -4$, range $(-4, \infty)$, $y$-intercept 2
Ejemplo resuelto: Decay and a flip
Find the asymptote and range of $y = -\left(\frac{1}{2}\right)^{x} + 5$.
- $\left(\frac{1}{2}\right)^x > 0$, so $-\left(\frac{1}{2}\right)^x < 0$ and $y < 5$.
Respuesta: Asymptote $y = 5$; range $(-\infty, 5)$
Error común
Moving the asymptote with a horizontal shift. In $y = 2^{x - 3}$ the asymptote is still $y = 0$: only the vertical shift $k$ moves it. Check by plugging in a very negative $x$, like $x = -20$: $2^{-23}$ is about 0.0000001.