Question 737564: Hyperbolas:
Graph each hyperbola. Identify the verices, foci, and asymptotes
X^2/4 - y^2/16 = 1
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! Graph each hyperbola. Identify the verices, foci, and asymptotes
This is a hyperbola that has a horizontal transverse axis (x-term listed first ahead of y-term)
Its standard form of equation: 
For given hyperbola:
center: (0,0)
a^2=4
a=2
vertices: (0±a,0)=(0±2,0)=(-2,0) and (2,0)
..
b^2=16
b=4
..
c^2=a^2+b^2=4+16=20
c=√20≈4.47
foci: (0±c,0)=(0±4.47,0)=(-4.47,0) and (4.47,0)
..
Asymptotes: (equations of lines that go thru center, y=mx+b, m=slope, b= y-intercept)
Slopes of asymptotes of hyperbolas with horizontal transverse axis=±b/a=±4/2=±2
..
Equation of asymptote with negative slope=-2
y=-2x+b
b=0
y=-2x
..
Equation of asymptote with positive slope=2
y=2x+b
b=0
y=2x
...
y=±(4x^2-16)^.5
See graph below as a visual check:
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