Question

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

Answer 1

This specific parabola should be transformed into the form

`y = (x - h)^2 + k`

To do that, we need to perform Completing the Square on `x`

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Remember that `(ax + b)^2 = a^2x^2 + 2abx + b^2`

For this problem, we have `a = 1` and `2ab = 3`

`=> 2(1)b = 3`
`=> b = 3/2`

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Hence, for `x` to be a perfect square, we need to add `b^2`.
However, we want both sides of the equation to remain equal, so we substract the same value again

`y = (x^2 + 3x + (3/2)^2) - 4 - (3/2)^2`

Simplifying, we have
`y = (x + 3/2)^2 - 25/4`

The parabola's vertex is at `(-3/2, 25/4)`.