What is the vertices (x^2}{16}-\frac{y^2)/9 =1 ?

The vertices (x^2}{16}-\frac{y^2)/9 =1 is (4,0),(-4,0)

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vertices (x^2}{16}-\frac{y^2)/9 =1

vertices

\frac{x ^{2}}{16} - \frac{y ^{2}}{9} = 1

(4, 0), (- 4, 0)

Solution steps

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

Calculate Hyperbola properties

\frac{x ^{2}}{16} - \frac{y ^{2}}{9} = 1 :

Right-left Hyperbola with

(h, k) = (0, 0), a = 4, b = 3

(0 + 4, 0), (0 - 4, 0)

Refine

(4, 0), (- 4, 0)

9 x^{2} - 16 y^{2} = 144

\frac{y ^{2}}{16} - \frac{x ^{2}}{9} = 1

x = 2 y^{2}

f (x) = x^{2} - 6 x + 1

x^{2} + \frac{y ^{2}}{64} = 1

What is the vertices (x^2}{16}-\frac{y^2)/9 =1 ?
The vertices (x^2}{16}-\frac{y^2)/9 =1 is (4,0),(-4,0)