# Graph -16x^2+96x-144+4y^2-32y=0 -16×2+96x-144+4y2-32y=0
Find the standard form of the hyperbola.
Add 144 to both sides of the equation.
-16×2+96x+4y2-32y=144
Complete the square for -16×2+96x.
Use the form ax2+bx+c, to find the values of a, b, and c.
a=-16,b=96,c=0
Consider the vertex form of a parabola.
a(x+d)2+e
Substitute the values of a and b into the formula d=b2a.
d=962(-16)
Simplify the right side.
Cancel the common factor of 96 and 2.
Factor 2 out of 96.
d=2⋅482⋅-16
Cancel the common factors.
Factor 2 out of 2⋅-16.
d=2⋅482(-16)
Cancel the common factor.
d=2⋅482⋅-16
Rewrite the expression.
d=48-16
d=48-16
d=48-16
Cancel the common factor of 48 and -16.
Factor 16 out of 48.
d=16(3)-16
Move the negative one from the denominator of 3-1.
d=-1⋅3
d=-1⋅3
Multiply -1 by 3.
d=-3
d=-3
Find the value of e using the formula e=c-b24a.
Simplify each term.
Raise 96 to the power of 2.
e=0-92164⋅-16
Multiply 4 by -16.
e=0-9216-64
Divide 9216 by -64.
e=0+144
Multiply -1 by -144.
e=0+144
e=0+144
e=144
e=144
Substitute the values of a, d, and e into the vertex form a(x+d)2+e.
-16(x-3)2+144
-16(x-3)2+144
Substitute -16(x-3)2+144 for -16×2+96x in the equation -16×2+96x+4y2-32y=144.
-16(x-3)2+144+4y2-32y=144
Move 144 to the right side of the equation by adding 144 to both sides.
-16(x-3)2+4y2-32y=144-144
Complete the square for 4y2-32y.
Use the form ax2+bx+c, to find the values of a, b, and c.
a=4,b=-32,c=0
Consider the vertex form of a parabola.
a(x+d)2+e
Substitute the values of a and b into the formula d=b2a.
d=-322(4)
Simplify the right side.
Cancel the common factor of 32 and 2.
Factor 2 out of 32.
d=-2⋅162⋅4
Cancel the common factors.
Factor 2 out of 2⋅4.
d=-2⋅162(4)
Cancel the common factor.
d=-2⋅162⋅4
Rewrite the expression.
d=-164
d=-164
d=-164
Cancel the common factor of 16 and 4.
Factor 4 out of 16.
d=-4⋅44
Cancel the common factors.
Factor 4 out of 4.
d=-4⋅44(1)
Cancel the common factor.
d=-4⋅44⋅1
Rewrite the expression.
d=-41
Divide 4 by 1.
d=-1⋅4
d=-1⋅4
d=-1⋅4
Multiply -1 by 4.
d=-4
d=-4
Find the value of e using the formula e=c-b24a.
Simplify each term.
Raise -32 to the power of 2.
e=0-10244⋅4
Multiply 4 by 4.
e=0-102416
Divide 1024 by 16.
e=0-1⋅64
Multiply -1 by 64.
e=0-64
e=0-64
Subtract 64 from 0.
e=-64
e=-64
Substitute the values of a, d, and e into the vertex form a(x+d)2+e.
4(y-4)2-64
4(y-4)2-64
Substitute 4(y-4)2-64 for 4y2-32y in the equation -16×2+96x+4y2-32y=144.
-16(x-3)2+4(y-4)2-64=144-144
Move -64 to the right side of the equation by adding 64 to both sides.
-16(x-3)2+4(y-4)2=144-144+64
Simplify 144-144+64.
Subtract 144 from 144.
-16(x-3)2+4(y-4)2=0+64
-16(x-3)2+4(y-4)2=64
-16(x-3)2+4(y-4)2=64
Divide each term by 64 to make the right side equal to one.
-16(x-3)264+4(y-4)264=6464
Simplify each term in the equation in order to set the right side equal to 1. The standard form of an ellipse or hyperbola requires the right side of the equation be 1.
(y-4)216-(x-3)24=1
(y-4)216-(x-3)24=1
This is the form of a hyperbola. Use this form to determine the values used to find vertices and asymptotes of the hyperbola.
(y-k)2a2-(x-h)2b2=1
Match the values in this hyperbola to those of the standard form. The variable h represents the x-offset from the origin, k represents the y-offset from origin, a.
a=4
b=2
k=4
h=3
The center of a hyperbola follows the form of (h,k). Substitute in the values of h and k.
(3,4)
Find c, the distance from the center to a focus.
Find the distance from the center to a focus of the hyperbola by using the following formula.
a2+b2
Substitute the values of a and b in the formula.
(4)2+(2)2
Simplify.
Raise 4 to the power of 2.
16+(2)2
Raise 2 to the power of 2.
16+4
20
Rewrite 20 as 22⋅5.
Factor 4 out of 20.
4(5)
Rewrite 4 as 22.
22⋅5
22⋅5
Pull terms out from under the radical.
25
25
25
Find the vertices.
The first vertex of a hyperbola can be found by adding a to k.
(h,k+a)
Substitute the known values of h, a, and k into the formula and simplify.
(3,8)
The second vertex of a hyperbola can be found by subtracting a from k.
(h,k-a)
Substitute the known values of h, a, and k into the formula and simplify.
(3,0)
The vertices of a hyperbola follow the form of (h,k±a). Hyperbolas have two vertices.
(3,8),(3,0)
(3,8),(3,0)
Find the foci.
The first focus of a hyperbola can be found by adding c to k.
(h,k+c)
Substitute the known values of h, c, and k into the formula and simplify.
(3,4+25)
The second focus of a hyperbola can be found by subtracting c from k.
(h,k-c)
Substitute the known values of h, c, and k into the formula and simplify.
(3,4-25)
The foci of a hyperbola follow the form of (h,k±a2+b2). Hyperbolas have two foci.
(3,4+25),(3,4-25)
(3,4+25),(3,4-25)
Find the eccentricity.
Find the eccentricity by using the following formula.
a2+b2a
Substitute the values of a and b into the formula.
(4)2+(2)24
Simplify.
Simplify the numerator.
Raise 4 to the power of 2.
16+(2)24
Raise 2 to the power of 2.
16+44
204
Rewrite 20 as 22⋅5.
Factor 4 out of 20.
4(5)4
Rewrite 4 as 22.
22⋅54
22⋅54
Pull terms out from under the radical.
254
254
Cancel the common factor of 2 and 4.
Factor 2 out of 25.
2(5)4
Cancel the common factors.
Factor 2 out of 4.
252⋅2
Cancel the common factor.
252⋅2
Rewrite the expression.
52
52
52
52
52
Find the focal parameter.
Find the value of the focal parameter of the hyperbola by using the following formula.
b2a2+b2
Substitute the values of b and a2+b2 in the formula.
2225
Simplify.
Cancel the common factor of 22 and 2.
Factor 2 out of 22.
2⋅225
Cancel the common factors.
Factor 2 out of 25.
2⋅22(5)
Cancel the common factor.
2⋅225
Rewrite the expression.
25
25
25
Multiply 25 by 55.
25⋅55
Combine and simplify the denominator.
Multiply 25 and 55.
2555
Raise 5 to the power of 1.
25515
Raise 5 to the power of 1.
255151
Use the power rule aman=am+n to combine exponents.
2551+1
2552
Rewrite 52 as 5.
Use axn=axn to rewrite 5 as 512.
25(512)2
Apply the power rule and multiply exponents, (am)n=amn.
25512⋅2
Combine 12 and 2.
25522
Cancel the common factor of 2.
Cancel the common factor.
25522
Divide 1 by 1.
2551
2551
Evaluate the exponent.
255
255
255
255
255
The asymptotes follow the form y=±a(x-h)b+k because this hyperbola opens up and down.
y=±2⋅(x-(3))+4
Simplify to find the first asymptote.
Remove parentheses.
y=2⋅(x-(3))+4
Simplify 2⋅(x-(3))+4.
Simplify each term.
Multiply -1 by 3.
y=2⋅(x-3)+4
Apply the distributive property.
y=2x+2⋅-3+4
Multiply 2 by -3.
y=2x-6+4
y=2x-6+4
y=2x-2
y=2x-2
y=2x-2
Simplify to find the second asymptote.
Remove parentheses.
y=-2⋅(x-(3))+4
Simplify -2⋅(x-(3))+4.
Simplify each term.
Multiply -1 by 3.
y=-2⋅(x-3)+4
Apply the distributive property.
y=-2x-2⋅-3+4
Multiply -2 by -3.
y=-2x+6+4
y=-2x+6+4
y=-2x+10
y=-2x+10
y=-2x+10
This hyperbola has two asymptotes.
y=2x-2,y=-2x+10
These values represent the important values for graphing and analyzing a hyperbola.
Center: (3,4)
Vertices: (3,8),(3,0)
Foci: (3,4+25),(3,4-25)
Eccentricity: 52
Focal Parameter: 255
Asymptotes: y=2x-2, y=-2x+10
Graph -16x^2+96x-144+4y^2-32y=0     