Lesson 4: Modeling with Quadratic Equations

Property

To find a parabola's equation from its vertex $(x_v, y_v)$ and another point $(x, y)$:

1. Substitute the vertex into the vertex form: $y = a(x - x_v)^2 + y_v$.
2. Substitute the coordinates of the other point for $x$ and $y$.
3. Solve for the value of $a$.
4. Write the final equation with the value of $a$.

Examples

• A parabola has a vertex at $(3, 4)$ and passes through $(5, 12)$. Start with $y = a(x - 3)^2 + 4$. Substitute the point: $12 = a(5 - 3)^2 + 4$, which gives $8 = 4a$, so $a=2$. The equation is $y = 2(x - 3)^2 + 4$.

• A ball's path has a vertex at $(8, 12)$ and starts at $(0, 4)$. Using $y = a(x - 8)^2 + 12$, we plug in $(0,4)$: $4 = a(0 - 8)^2 + 12$. This gives $-8 = 64a$, so $a = -\frac{1}{8}$. The equation is $y = -\frac{1}{8}(x - 8)^2 + 12$.

Property

The intercept form of a quadratic equation is

Examples

Property

A quadratic equation can be written in the form

Examples

• To find the parabola through $(1, 3)$, $(3, 5)$, and $(4, 9)$, we solve the system: $a + b + c = 3$, $9a + 3b + c = 5$, and $16a + 4b + c = 9$. The solution is $a=1, b=-3, c=5$, so the equation is $y = x^2 - 3x + 5$.

• A parabola passes through $(0, 8)$, $(1, 5)$, and $(2, 6)$. The system is $c = 8$, $a+b+c=5$, and $4a+2b+c=6$. Substituting $c=8$ gives $a+b=-3$ and $4a+2b=-2$. The solution is $a=2, b=-5, c=8$, so $y = 2x^2 - 5x + 8$.