Question

What is the the vertex of `y = (x -3)^2+4x-5`?

Answer 1

The solution set(or vertex set) is: `S = {-5,-21}.`

Explanation:

The standard formula of the quadratic function is:
`y = Ax^2 + Bx +C`

`(x-3)^2` is a notable product, so do this:
Square the first number -(signal inside the parenthesis) 2 * first number * second number + second number squared
`x^2 - 6x + 9`

Now, substitute it the main equation:
`y = x^2 - 6x + 9 +4x - 5 = x^2 +10x +4`, so
`y = x^2 +10x +4` `to` Now, it agrees with the standard formula.

To find the point of the vertex in `x` axis, we apply this formula:
`x_(vertex) = -b/(2a) = -10/2 = -5`

To find the point of the vertex in `y` axis, we apply this formula:
`y_(vertex) = - triangle/(4a) = - (b^2 - 4ac)/(4a) = -(100 -4 * 1 * 4)/4 = -21`

Then, the solution set(or vertex set) is: `S = {-5,-21}.`