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

The vertices (x^2)/9-(y^2)/(16)=1 is (3,0),(-3,0)

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vertices (x^2)/9-(y^2)/(16)=1

vertices

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

(3, 0), (- 3, 0)

Solution steps

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

Calculate Hyperbola properties

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

Right-left Hyperbola with

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

(0 + 3, 0), (0 - 3, 0)

Refine

(3, 0), (- 3, 0)

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

f (x) = x^{2} + 4 x - 1

(x - h)^{2} = 4 p (y - k)

x^{2} + (y - 1)^{2} = 1

x = y^{2} + 1

What is the vertices (x^2)/9-(y^2)/(16)=1 ?
The vertices (x^2)/9-(y^2)/(16)=1 is (3,0),(-3,0)