Question

The parabolic curve of a certain camera lens can be represented by the equation `y=10x^2+50x+63.2`. What are the coordinates of the vertex?

Answer 1

The vertex is, `(5/2, 7/10)`

Explanation:

The vertex form of a parabola of this type is:

`y = a(x - h)^2 + k" [1]"`

where "a" is the leading coefficient of the `x^2` term and `(h,k)` is the vertex.

Expand the square in equation [1]:

`y = a(x^2 - 2hx + h^2) + k" [2]"`

To help us put the given equation into the form of equation [2], add zero to the given equation in the form of `ah^2 - ah^2` where `a = 10`:

`y = 10x^2 - 50x + 10h^2 - 10h^2 + 63.2`

NOTE: I used `ah^2 - ah^2` so that it would be easy to factor.

Factor a 10 out of the first 3 terms:

`y = 10(x^2 - 5x + h^2) - 10h^2 + 63.2" [3]"`

Set the second term in in equation [2] equal to the second term in equation [3] and then solve for h:

`-2hx = -5x`

`h = 5/2`

Substitute the square in equation [1] into equation [3]:

`y = 10(x - h)^2 - 10h^2 + 63.2" [4]"`

Substitute `5/2` for every h in equation [4]:

`y = 10(x - 5/2)^2 - 10(5/2)^2 + 63.2" [5]"`

Simplify the constant term:

`y = 10(x - 5/2)^2 + 7/10" [6]"`

The vertex, `(5/2, 7/10)`, is found by observation.