Problema Solution
Hi can you please help me with this question, graph the equation equal to: 2x^2-2x-2y-1=0 and find the vertix, focus point and axis of symmetry. Thank you all your help...
Answer provided by our tutors
2x^2-2x-2y-1=0.
2x^2-2x-1=2y.
y=(2x^2-2x-1)/2.
=x^2-x-1/2.
vertix, focus point and axis of symmetry.
vertex : ( -b/2a, f(-b/2a).
-b/2a =1/2.
f(1/2)=(1/2)^2-1/2-1/2.
=1/4-1 =-3/4.
vertex (1/2,-3/4).
focus point (-b/2a, c-(b^2-1)/4a).
c-(b^2-1)/4a =-1/2-(1-1)/4*1 =-1.
focus (1/2,-1).
axis of symmetry is x=-b/2a
x=1/2 or 2x-1=0.