Question

What is the the vertex of `y=x^2+15x-30`?

Answer 1

I found: `(-7.5,-86.25)`

Explanation:

There are two ways to find the coordinates of the vertex:

1) knowing that the `x` coordinate is given as:
`x_v=-b/(2a)` and considering your function in the general form:
`y=ax^2+bx+c`;

in your case:
`a=1`
`b=15`
`c=-30`
so:
`x_v=-15/(2)=-7.5`
by substituting this value into your original equation you get the corresponding `y_v` value:
`y_v=(-15/2)^2+15(-15/2)-30=(225-450-120)/4=-345/4=-86.25`

2) usig the derivative (but I am not sure you know this procedure):
Derive your function:
`y'=2x+15`
set it equal to zero (to find the point of zero slope...the vertex):
`y'=0`
i.e.
`2x+15=0`
and solve to get:
`x=-15/2` as before!

Graphically:
graph{x^2+15x-30 [-240.5, 240.3, -120.3, 120.3]}