Question

What is the vertex of `y=x^2-2x-35`?

Answer 1

`color(blue)("Vertex" -> (x,y)-> (1,-36)`

I have shown a really 'cool' trick to solve this.

Explanation:

To demonstrate how useful the method I am about to show you is.

Just by looking at the given equation I determine that `x_("vertex")` is at `x=+1`

Then it is just a matter of substitution to find `y_("vertex")` which is -36
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Ok! lets deal with the method!

The standardised equation structure `ax^2+bx+c`

In you case `a=1`

Change `ax^2+bx+c" to " a(x^2+b/a x) +c`

Then `x_("vertex") = (-1/2)xx b/a`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Solving your question")`

`a=1`

`b/a = (-2)/1=-2`

`x_("vertex")=(-1/2)xx(-2) = color(red)(+1)`

`color(brown)("The above is part way to developing the vertex equation format")`

Sometimes the number are a bit more difficult to work out.

`y_("vertex")=(color(red)(1))^2-2(color(red)(1))-35`

`y_("vertex")=1-2-35=-36`

`color(blue)("Vertex" -> (x,y)-> (1,-36)`