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.

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: [tex]x=h[/tex]

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: [tex]x=6[/tex] 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: [tex](6, 6)[/tex]. Using this we can plug in some values into the vertex form: [tex]y=a(x-6)^2+6[/tex], 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: [tex](7, 5)[/tex] into the vertex form as (x, y) to solve for the "a" value, so let's do that.

Original Equation:

[tex]y=a(x-6)^2+6[/tex]

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

[tex]5=a(7-6)^2+6[/tex]

Simplify inside parenthesis:

[tex]5=a(1)^2+6[/tex]

Square the one:

[tex]5=a+6[/tex]

Subtract 6 from both sides:

[tex]-1=a[/tex]

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