Esta lección aún no está traducida. Por ahora se muestra en inglés; la versión en español está en camino.
To write an exponential model you need a starting value and a factor. A story may give both, as in growth and decay rates, or you may have to find them from data points, as with the ratio of a geometric sequence.
💡 Por qué funciona
In $y = a \cdot b^x$, the point at $x = 0$ gives $a$ directly, because $b^0 = 1$. A second point then pins down $b$: if $y$ goes from $a$ at $x = 0$ to $y_2$ at $x = 2$, then $a \cdot b^2 = y_2$, so $b^2 = \frac{y_2}{a}$. In data with equal steps, the factor shows up as the constant ratio of neighboring values. If the differences are constant instead, a line fits better.
🔑 Idea clave
Find $a$ (the value at $x = 0$) and $b$ (the factor per step, from a ratio of outputs), then write $y = a \cdot b^x$. Check that the model reproduces the data.
Set $a$ and $b$ so the table matches the data in the first example: $5, 15, 45, 135$.
Worked examples
✏️ Ejemplo resuelto: From two points
An exponential function passes through $(0, 5)$ and $(2, 45)$. Write it.
$x = 0$ gives $a = 5$.
$5 \cdot b^2 = 45$, so $b^2 = 9$ and $b = 3$ ($b$ must be positive).
Check: $5 \cdot 3^1 = 15$ would be the value at $x = 1$, and $5 \cdot 3^2 = 45$.
Respuesta: $y = 5 \cdot 3^x$
✏️ Ejemplo resuelto: From a story
A sample of $400$ mg of a substance loses half its mass every $6$ hours. Write a model.
Every $6$ hours the mass is multiplied by $\frac{1}{2}$.
After $t$ hours there have been $\frac{t}{6}$ half-lives.
Fitting a linear model to data that changes by a constant ratio. Data like $5, 15, 45, 135$ has differences $10, 30, 90$ that keep growing, so a straight line will miss badly after a few steps. Always compute both differences and ratios before choosing.
Práctica
Apruebas con 3 respuestas correctas, las 2 últimas seguidas. Cada respuesta correcta da 3 XP.