Question

find the center, transverse axis, vertices, foci, and asymptotes. graph the equation.
\\( \frac { x ^ { 2 } } { 16 } - \frac { y ^ { 2 } } { 9 } = 1 \\)
the center is at \\( ( 0,0 ) \\).
(type an ordered pair. type exact answers for each coordinate, using radicals as needed.)
the transverse axis is along the \\( x \\)-axis.
the vertices are at \\( ( - 4,0 ), ( 4,0 ) \\)
(simplify your answer. type ordered pairs. type exact answers, using radicals as needed. use a comma to separate
answers as needed.)
the foci are at \\( ( - 5,0 ), ( 5,0 ) \\).
(simplify your answer. type ordered pairs. type exact answers, using radicals as needed. use a comma to separate
answers as needed.)
the asymptote with positive slope is \\( \square \\), and the asymptote with negative slope is \\( \square \\)
(simplify your answers. type equations. use integers or fractions for any numbers in the equations.)

Explanation:

Step1: Recall the standard form of hyperbola

The standard form of a hyperbola with center \((h,k)\) and transverse axis along the \(x -\)axis is \(\frac{(x - h)^{2}}{a^{2}}-\frac{(y - k)^{2}}{b^{2}}=1\). For the given equation \(\frac{x^{2}}{16}-\frac{y^{2}}{9}=1\), we have \(h = 0,k = 0,a^{2}=16\), so \(a = 4\), and \(b^{2}=9\), so \(b = 3\).

Step2: Find the equations of asymptotes

The equations of the asymptotes for a hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\) (transverse axis along \(x -\)axis) are \(y=\pm\frac{b}{a}x\).
Substituting \(a = 4\) and \(b = 3\) into the formula for the asymptotes:
For the positive - slope asymptote: \(y=\frac{b}{a}x=\frac{3}{4}x\)
For the negative - slope asymptote: \(y=-\frac{b}{a}x=-\frac{3}{4}x\)

Answer:

The asymptote with positive slope is \(y = \frac{3}{4}x\), and the asymptote with negative slope is \(y=-\frac{3}{4}x\)