# Pre-Calculus

posted by .

Find the coordinates of the center, the foci, the vertices, and the equations of the asymptotes of the graph of the equation: (x+1)^2/2-y^2/8=1.

Center: (-1,0)
foci:(-1+/-sqrt1, 0)
vertices: (-1+/-sqrt2, 0)
equations of the asymptotes: y-0=sqrt8/sqrt2(x+1)

Is this correct and if not, how do I figure this out??

• Pre-Calculus -

hyperbola due to - sign
yes, center at -1,0 because of(x+1)^2 and (y)^2
a =sqrt 2
b = sqrt 8 = 2 sqrt 2
so vertices at y =0 x = -1+sqrt 2 and -1 -sqrt 2
center to focus = sqrt(2+8) = sqrt 10
so at y = 0 and x = -1 +/- sqrt 10
slope of asymptotes = +/- b/a = sqrt2
so
y = (+/- sqrt 2) - 1
so yes, correct

## Similar Questions

1. ### Pre-Calculus

I am having trouble with several problems: 1) Find the coordinates of the center, foci, and the vertices, and the equations of the asymptotes of the graph of the equation (x+1)^2/2 - y^2/8 =1 2) Write the standard form of the equation …
2. ### Pre-Calculus

I am having trouble with several problems: 1) Find the coordinates of the center, foci, and the vertices, and the equations of the asymptotes of the graph of the equation (x+1)^2/2 - y^2/8 =1 2) Write the standard form of the equation …

Identify the graph of the equation 4x^2-25y^2=100. Then write the equation of the translated graph for T(5,-2) in general form. Answer: hyperbola; 4(x-5)^2 -25(y+2)^2=100 2)Find the coordinates of the center, the foci, and the vertices, …
4. ### Pre-calculus

For the ellipse with equation 5x^2+64y^2+30x+128y-211=0, find the cooridinates of the center, foci, and vertices. Then, graph the equation. my answer is: coordinates of center: (-3,1) foci:(-11,-1) and (4.7,-1) vertices: (-11,-1) (5,-1) …
5. ### Algebra II-Please check my calculation

Find center, vertices, co-vertices and foci for the following; (x-3(^2/49 + (y-4)^2/4 = 1 Center would be (3,4) A=7 b=2 Vertices would be found this way: (h+-a,k) (3-7,4) (3+7,4) Vertices = (-4,4) (10,4) Co-vertices (h,k+-b) (3,4-2) …
6. ### Math

Find the vertices, center, asymptotes, and foci of the ellipse and sketch its graph: 9x^2-4y^2-18x+8y+41=0
7. ### algebra 2

Graph the equation. Identify the vertices, foci, and asymptotes of the hyperbola. Y^2/16-x^2/36=1 I think....vertices (0,4),(0,-4) foci?
8. ### Pre-calc

Heres the equation: (x^2)-(4y^2)-(4x)+(24y)-(36)=0 Your supposed to find the coordinates of the center, foci, and vertices, and the equations of the asymptotes of its graph. So first you complete the square right?
9. ### precal

complete the square to identify what type of conic you have, identify the key parts indicated and then graph conic. parabola: vertex,focus, directrix, focal diameter. ellipse: center,vertices, foci, eccentricity. hyperbola: center, …
10. ### math

Identify the direction of opening,the coordinates of the center, the vertices, and the foci. Find the equations of the asymptotes and sketch the graph. x^2/16-y^2/9=1

More Similar Questions