Given f(x)=x^2-4x+5, what's the value when f(2)
1 answer:
You might be interested in
Hello,
Vertices are on a line parallele at ox (y=-3)
The hyperbola is horizontal.
Equation is (x-h)²/a²- (y-k)²/b²=1
Center =middle of the vertices=((-2+6)/2,-3)=(2,-3)
(h+a,k) = (6,-3)
(h-a,k)=(-2,-3)
==>k=-3 and 2h=4 ==>h=2
==>a=6-h=6-2=4 (semi-transverse axis)
Foci: (h+c,k) ,(h-c,k)
h=2 ==>c=8-2=6
c²=a²+b²==>b²=36-4²=20
Equation is:
Vertices are on a line parallele at ox (y=-3)
The hyperbola is horizontal.
Equation is (x-h)²/a²- (y-k)²/b²=1
Center =middle of the vertices=((-2+6)/2,-3)=(2,-3)
(h+a,k) = (6,-3)
(h-a,k)=(-2,-3)
==>k=-3 and 2h=4 ==>h=2
==>a=6-h=6-2=4 (semi-transverse axis)
Foci: (h+c,k) ,(h-c,k)
h=2 ==>c=8-2=6
c²=a²+b²==>b²=36-4²=20
Equation is: