Esta lección aún no está traducida. Por ahora se muestra en inglés; la versión en español está en camino.
Two points determine a line. For a parabola you need three pieces of information, because $y = ax^2 + bx + c$ has three unknown numbers. Which form you start from depends on what you are given, and the vertex and the zeros are the most useful facts to have.
🔑 Idea clave
Given the vertex $(h, k)$, start from $y = a(x - h)^2 + k$. Given the zeros $r$ and $s$, start from $y = a(x - r)(x - s)$. Either way, one more point pins down $a$. Given three ordinary points, substitute each into $y = ax^2 + bx + c$ and solve the system.
💡 Por qué funciona
Each form already has the given facts built in. In vertex form, every value of $a$ gives a parabola with vertex $(h, k)$; the extra point picks out the one stretch that passes through it. In factored form, every value of $a$ gives zeros at $r$ and $s$, because a product is 0 when a factor is 0. With three plain points, each one turns into a true equation in $a$, $b$ and $c$, and three equations are enough to find three unknowns.
Factored-form sliders: $r$ and $s$ are the zeros. Set the zeros to $-1$ and $3$, then change $a$ until the vertex is $(1, -8)$.
Worked examples
✏️ Ejemplo resuelto: From the vertex and a point
A parabola has vertex $(2, -3)$ and passes through $(4, 5)$. Find its equation.
$y = a(x - 2)^2 - 3$.
Substitute $(4, 5)$: $5 = a(2)^2 - 3$, so $4a = 8$ and $a = 2$.
Respuesta: $y = 2(x - 2)^2 - 3$
✏️ Ejemplo resuelto: From the zeros and a point
A parabola has zeros $-1$ and $5$ and $y$-intercept $10$. Find its equation.
$y = a(x + 1)(x - 5)$.
The $y$-intercept is the point $(0, 10)$: $10 = a(1)(-5) = -5a$, so $a = -2$.
Respuesta: $y = -2(x + 1)(x - 5)$
✏️ Ejemplo resuelto: From three points
Find $y = ax^2 + bx + c$ through $(0, 1)$, $(1, 4)$ and $(2, 11)$.
$(0, 1)$ gives $c = 1$.
$(1, 4)$ gives $a + b + 1 = 4$, so $a + b = 3$.
$(2, 11)$ gives $4a + 2b + 1 = 11$, so $4a + 2b = 10$, or $2a + b = 5$.
Subtract the first from the second: $a = 2$, then $b = 1$.
Respuesta: $y = 2x^2 + x + 1$
⚠️ Error común
Writing the factor for the zero $-1$ as $(x - 1)$. A zero at $x = r$ makes the factor $(x - r)$ equal 0, so the zero $-1$ gives $(x + 1)$. Check by substituting: $(-1) + 1 = 0$. Also check your final equation against every given point.
Práctica
Apruebas con 3 respuestas correctas, las 2 últimas seguidas. Cada respuesta correcta da 2 XP.