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

1 Answer
Gió
Jun 3, 2016

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]}

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