find the vertex of the hyperbola help!!!!!!?

Question

find the vertex of the hyperbola
y^2=2x^2-10x+3

Answer 1

This is the equation of a hyperbola, not a parabola. This has vertices when y = 0, ie. when

2 x^2 - 10x + 3 = 0.

Solve this quadratic, and then, if the roots are a and b, you vertices are at (a, 0) and (b, 0).

Answer 2

Hyperbolas come in pairs and therefore have two vertices.

y² = 2x² - 10x + 3
y² - 2x² + 10x = 3

Complete the square for x.

y² - 2(x² - 5x) = 3
y² - 2(x² - 5x + 25/4) = 3 - 2*(25/4)
y² - 2(x - 5/2)² = -19/2

2(x - 5/2)² - y² = 19/2
(x - 5/2)²/(19/4) - y²/(19/2) = 1

a² = 19/4
a = √19/2

The center of the hyperbola (h,k) = (5/2, 0).

Since the x² term is positive the hyperbolas open sideways. The vertices are

(h - a, k) and (h + a, k)
or
[(5 - √19)/2, 0] and [(5 + √19)/2, 0]

Answer 3

you want a y^2 - (x-h)^2 = 1 type format..

y^2 = 2(x^2 - 5x + 25/4) + 3 - 50/4

y^2 -2(x-5)^2 = -19/2

-2/19 * y^2 + 4/19 * (x-5)^2 = 1

Line of symmetry: x = 5

Hyperbola opens left-right

Vertices: (5 +/- sqrt(19)/2, 0)

Answer 4

Download a useful program called Graphmatica.
It is great for any function its amazing, sketches, finds integrals derivatives, sketches derivatives and all functions. Its very useful if im checking an answer.

Answer 5

You said that was a parabola.

Answer 6

(-5,0)