.
#y=2x^2-4x-30#
This is a quadratic equation, i.e. the equation of a parabola. Let's find its vertex, #x#-intercepts, and #y#-intercept. This is in the forms of:
#y=ax^2+bx+c#
From this form, we can find the #x# coordinate of the vertex by finding the value of #(-b)/(2a)#:
#(-b)/(2a)=(-(-4))/(2(2))=4/4=1#
We then plug this into the equation to find the #y# coordinate of the vertex:
#y=2(1)^2-4(1)-30=2-4-30=-32#
So, vertex #(1,-32)#
Now, let's find the #x#-intercepts by setting the equation equal to zero and solving for the roots:
#2x^2-4x-30=0#
#2(x^2-2x-15)=0#
#x^2-2x-15=0#
#(x-5)(x+3)=0#
#x=5, -3#
As such, the #x#-intercepts are #(-3,0)# and #(5,0)#
The #y#-intercept can be found by setting #x=0#
#y=-30#
#y#-intercept is #(0,-30#
The graph of it is:
graph{2x^2-4x-30 [-80, 80, -40, 40]}
The domain of the function is all values of #x#:
Domain: #-oo< x < oo#
Range is from the #y# value of the vertex to infinity:
Range: #-32 < y < oo#
