A hyperbola is represented by the equation x^2/9 - (y - 1)^2 = 1. Identify the asymptotes of the graph.
Out of the options I'm given, it's either these two:
y = (sqrt5/3) x + 1 and y = -(sqrt5/5) x + 1
y = (sqrt5/3) (x + 1) and y = -(sqrt5/5) (x + 1)
The answer I keep coming up with is y = (xsqrt5/3) + 1 and y = -(xsqrt5/5) + 1, so I don't know where I'm wrong and what option to pick.
y= sqrt5/3 x+1 and y= -sqrt5/3 x+1
no parentheses
if you’re doing the comic sections test i got
1. c 2sqrt10
2. ellipse
3. a. (0,-4) r=sqrt 3
4. b (x+3)^2 +(y-2)^2=9
5. d (x-4)^2+(y-3)^2=4
6. a. (x-3)^2/16 + (y-1)^2/4=1
7. c. (x+4)^2/4 +(y+3)^2/16 =1 (-4,-3 +- 2 sqrt3)
8. ellipse is vertical because a<b
9. a 152 million
10. c x^2/3
11. d (y-5)^2= 3(x-1)
12. c focus (4,3)
13. i got this one wrong so don’t pick 4
14. graph b should look like ) • ( where the center dot is on (5,2) and the other dots are at (1,2) and (9,2)
15. a (y+5)^2/9 - (x-2)^2/3 =1
16. a y= sqrt5/3 x+1 and y= -sqrt5/3 x+1
good luck
Moony has some wrong answers due to having different q's or typing wrong. Thus, the correct answers for those have questions like me
1. 2sqrt13
2. ellipse
3. C (0,-3) r=sqrt5
4. C x+1^2 y-4^2=1
5. D x-4 y+3=4
6. A x-3/16 + y-1/4
7. C (x+4)^2/4 +(y+3)^2/16 =1 (-4,-3 +- 2 sqrt3)
8. vertical
9. Ans: 152 If the sun is located at one focus and its coordinates are left-parenthesis 0,0 right-parenthesis, find Earth's farthest distance from the sun in millions of kilometers.
10. D x^2/3
11. C y-5=-3(x-1
12. focus 2,1 dir=6
13. Ans:4
A parabolic microphone used on the sidelines of a professional football game uses a reflective dish 16 in. wide and 4 in. deep. How far from the bottom of the dish should the microphone be placed?
14. Graph B as moony has mentioned 5,2 center
15. B y+3/25 - x-1/25 =1
16. A y= sqrt5/3 x+1 and y= -sqrt5/3 x+1
Well, it seems like you're getting a little tangled up in those options! Don't worry, I'm here to help untangle things for you.
To find the asymptotes of a hyperbola, we first need to determine the slope of these lines.
Let's focus on the x-coordinate terms in the given equation: x^2/9. We can rewrite this as (x/3)^2.
Now, let's compare that to the standard form of a hyperbola equation: (x-h)^2/a^2 - (y-k)^2/b^2 = 1.
Comparing the two equations, we can see that the center of the hyperbola is at (h, k) = (0, 1).
Since the x-coordinate term (x/3)^2 is positive, the hyperbola opens horizontally. The general form of the asymptotes is given by:
y = (±b/a)(x-h) + k.
In this case, since a^2 = 9, we have a = 3. Since b^2 = 1, we have b = 1. Plugging these values into the general form of the asymptotes, we get:
y = (±1/3)x + 1.
So, the correct answer is y = (±1/3)x + 1. That means none of the options you were given match up with the correct answer. It's okay to be wrong sometimes, it helps us learn and grow! Keep going and don't be discouraged.
To determine the asymptotes of a hyperbola, you need to consider the equation of the hyperbola and its general form. The general equation of a hyperbola is of the form:
[(x - h)^2 / a^2] - [(y - k)^2 / b^2] = 1
In your provided equation x^2/9 - (y - 1)^2 = 1, you can observe that the center of the hyperbola is at the point (h, k) = (0, 1) and the denominators are:
a^2 = 9, which means a = 3
b^2 = 1, which means b = 1
The asymptotes of a hyperbola are given by the equations:
y - k = ± (b / a)(x - h)
Substituting the values, you can determine the slope of the asymptotes:
y - 1 = ± (1 / 3)(x - 0)
Simplifying, you get:
y - 1 = ± (1/3)x
Now, we can rewrite this equation in the form y = mx + b:
y = (1/3)x + 1
and
y = -(1/3)x + 1
Comparing these equations to the options you have:
y = (sqrt5/3) x + 1 and y = -(sqrt5/5) x + 1
y = (sqrt5/3) (x + 1) and y = -(sqrt5/5) (x + 1)
None of the provided options match the correct equations for the asymptotes of the given hyperbola. The correct asymptotes for the hyperbola x^2/9 - (y - 1)^2 = 1 are:
y = (1/3)x + 1
and
y = -(1/3)x + 1
Your hyperbola has centre (0,1) , a = 3, and b = 1
vertices would be (3,1) and (-3,1)
To sketch this hyperbola I would faintly sketch a rectangle with corners at (3,2), (-3,2) , (-3,0) and (3,0)
One asymptote would run through (3,2) and (-3,0) and of course (0,1)
slope = 1/3
y = (1/3)x + 1 , since (0,1) is the y-intercept
similarly the other one would be y = (-1/3)x + 1
verification:
http://www.wolframalpha.com/input/?i=plot+x%5E2%2F9+-+(y+-+1)%5E2+%3D+1,+y+%3D+(1%2F3)x+%2B+1,+y+%3D+(-1%2F3)x+%2B+1
I have no idea where the square roots come from in your choices.