Question

Suppose a parabola has an axis of symmetry at x = 6, a maximum height of 6 and also passes through the point (7, 5). Write the equation of the parabola in vertex form.

Answer 1

`y=-(x-6)^2+6`

Explanation:

Vertex Form:

Vertex form of a parabola is expressed as:
`y=a(x-h)^2+k`

• (h, k) = vertex
• "a" determines vertical stretch/compression and direction

It's also important to note that the axis of symmetry is a vertical line which passes through vertex, so in this form, that axis of symmetry can be defined as such:
`x=h`

So if we're given the axis of symmetry we know the x-coordinate of the vertex.

Solving the Problem:

Since we're given the axis of symmetry to be:
`x=6` then that means the x-coordinate of the axis of symmetry is six. Another important thing to note about the vertex is it's y-value is either a minimum or maximum depending on whether it's opening up or down. So when we're given the maximum height to be 6, that's the y-coordinate of the vertex.

So now we know the coordinates to the vertex:
`(6, 6)`. Using this we can plug in some values into the vertex form:
`y=a(x-6)^2+6`, but we still don't know the "a" value in front (besides that it's negative). Luckily we're given a point on the equation! We can plug in the coordinates given:
`(7, 5)` into the vertex form as (x, y) to solve for the "a" value, so let's do that.

Original Equation:

`y=a(x-6)^2+6`

Passes through the point (7, 5), so substitute it for (x, y)

`5=a(7-6)^2+6`

Simplify inside parenthesis:

`5=a(1)^2+6`

Square the one:

`5=a+6`

Subtract 6 from both sides:

`-1=a`

So now let's plug this back into our equation:
`y=-1(x-6)^2+6`, although we don't have to explicitly put the one, and just put the negative sign which does the same thing:
`y=-(x-6)^2+6`