What is the vertices (x^2)/4-(y^2)/(16)=1 ?

The vertices (x^2)/4-(y^2)/(16)=1 is (2,0),(-2,0)

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vertices (x^2)/4-(y^2)/(16)=1

vertices

\frac{x ^{2}}{4} - \frac{y ^{2}}{16} = 1

(2, 0), (- 2, 0)

Solution steps

(h + a, k), (h - a, k)

Calculate Hyperbola properties

\frac{x ^{2}}{4} - \frac{y ^{2}}{16} = 1 :

Right-left Hyperbola with

(h, k) = (0, 0), a = 2, b = 4

(0 + 2, 0), (0 - 2, 0)

Refine

(2, 0), (- 2, 0)

f (x) = x^{2} + 4 x - 5

x^{2} + 9 y^{2} + 6 x - 90 y + 225 = 0

\frac{x ^{2}}{25} - \frac{y ^{2}}{9} = 1

\frac{x ^{2}}{49} + \frac{y ^{2}}{9} = 1

y^{2} - x^{2} = 1

What is the vertices (x^2)/4-(y^2)/(16)=1 ?
The vertices (x^2)/4-(y^2)/(16)=1 is (2,0),(-2,0)