Question

What is the vertex form of `x = (5y +3 )^2 + 2x+1`?

Answer 1

The vertex form for a parabola of this type is:

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

where `(h,k)` is the vertex and "a" is the leading coefficient.

Explanation:

Given: `x = (5y +3 )^2 + 2x+1`

Here is a graph of the original equation:

graph{x = (5y +3 )^2 + 2x+1 [-10, 10, -5, 5]}

Factor out a 5 from within the square:

`x = (5(y +3/5) )^2 + 2x+1`

Use the rules of squaring:

`x = (5)^2(y +3/5)^2 + 2x+1`

`x = 25(y +3/5)^2 + 2x+1`

The sign within the square must be negative:

`x = 25(y - (-3/5))^2 + 2x+1`

Subtract `2x` from both sides:

`-x = 25(y - (-3/5))^2+1`

Multiply both sides by -1:

`x = -25(y - (-3/5))^2-1`

The is the same form as equation [1] where `a = -25, k = -3/5 and h = -1`

Here is a graph of the vertex form:

graph{x = -25(y - (-3/5))^2-1 [-10, 10, -5, 5]}

Please observe that the graphs are identical.