# Thread: Writing equations from pieces of info

1. ## Writing equations from pieces of info

Given:
Direction opens up with minimum at -2
Passes through (0,0) and (2,0)

Write the equation of the parabola.

Let y = a(x - p)² + q

The q value of vertex is -2, so

y = a(x - p)² - 2

What do I do next? Pass through the points? How to solve this?

Thanks.

2. Originally Posted by shenton
Given:
Direction opens up with minimum at -2
Passes through (0,0) and (2,0)

Write the equation of the parabola.

Let y = a(x - p)² + q

The q value of vertex is -2, so

y = a(x - p)² - 2

What do I do next? Pass through the points? How to solve this?

Thanks.
Yes.

For (0, 0):
$\displaystyle 0 = a(-p)^2 - 2$

For (2, 0):
$\displaystyle 0 = a(2 - p)^2 - 2$

The first says $\displaystyle ap^2 = 2$ and the second says $\displaystyle ap^2 - 4ap + 4a - 2 = 0$.

Putting $\displaystyle ap^2 = 2$ into the second equation gives:
$\displaystyle -4ap + 4a = 4a(-p + 1) = 0$

so either a = 0 (a contradiction, if this were true then we won't have a parabola!) or $\displaystyle -p + 1 = 0$

Thus p = 1.

Thus $\displaystyle ap^2 = a = 2$ gives a = 2.

$\displaystyle y = 2( x - 1)^2 - 2$

-Dan

3. Hello, shenton!

Given: opens up with minimum at -2
Passes through (0,0) and (2,0)

Write the equation of the parabola.

The general equation of a "vertical" parabola is: .$\displaystyle y \:=\:ax^2 + bcx + c$

We have three points: .$\displaystyle (0,0),\;(2,0)$ . . . and the vertex: $\displaystyle (1,-2)$
. . If you plot the two points, you see they are x-intercepts.
. . The axis of symmetry lies halfway between them at $\displaystyle x = 1$
. . And that is how I located the vertex.

Substitute the points into the general equation.

$\displaystyle \begin{array}{ccc}(0,0): \\ (2,0): \\ (1,\text{-}2):\end{array} \begin{array}{ccc}0 \;= & a(0^2) + b(0) + c \\ 0 \:= & a(2^2) + b(2) + c \\ \text{-}2 \:= & a(1^2) + b(1) + c \end{array}\begin{array}{ccc}\Rightarrow\\ \Rightarrow\\ \Rightarrow\end{array}\begin{array}{ccc} \boxed{c \:=\:0} \\ 4a + 2b \:=\:0 \\ a + b\:=\:\text{-}2\end{array}\begin{array}{ccc}(1)\\(2)\\(3)\end{a rray}$

Divide (2) by 2: .$\displaystyle 2a + b \:=\:0$
. . Subtract (3): .$\displaystyle \text{-}a - b \:=\:2$
. .And we have: .$\displaystyle \boxed{ a = 2}$

Substitute into (3): .$\displaystyle 2 + b\:=\:\text{-}2\quad\Rightarrow\quad\boxed{b = \text{-}4}$

Therefore, the equation of the parabola is: .$\displaystyle y \:=\:2x^2 - 4x$