Question

Question: write an equation for each parabola with vertex at the origin

Problem: through (3,2); symmetric with respect to the x axis

Answer 1

Explanation:

Since the axis of symmetry is the x axis, we have a sideways parabola,

`{y}^(2) = x`

Since the vertex is at the orgin, our answer will be

`a(y - k) {}^(2) + h = 0`

`a {y}^(2) = x`

It passes through (3,2).

`a(2) {}^(2) = 3`

`a(4) = 3`

`a = (3)/(4)`

So our function is

`(3)/(4) {y}^(2) = x`

Answer 2

`x=\frac34y^2`

Explanation:

A graph with x-axis symmetry means that for every (x, y) point there is also a (x, -y) point on the graph, i.e. the x-axis acts like a mirror. Therefore, it is a sideways parabola.

The general form of this type of equation is
`x=ay^2+by+c`

Since we know that the curve has a vertex of (0, 0) then
`c=0`

Therefore,
`x=ay^2+by`

Given that (3, 2) is on the curve, then (3, -2) is also on the curve.

Substituting the given point (3, 2):

`3=a(2)^2+b(2)`

`3=4a+2b`

Substituting the given point (3, -2)

`3=a(-2)^2+b(-2)`

`3=4a-2b`

Adding the equations together to eliminate
`b`:

`\implies 6=8a`

`\implies 3=4a`

`\implies a=\frac34`

Therefore,

`x=\frac34y^2+by`

Again, substituting point (3, 2) means that b = 0

Therefore, the final equation is:

`x=\frac34y^2`