Intermediate Algebra: Functions and Graphs

The graph is a hyperbola with center at the origin. The \(y^2\)-term is positive, so the branches of the hyperbola open upward and downward. Because \(a^2=4\) and \(b^2=9\text{,}\) we have \(a=2\) and \(b=3\text{,}\) and the vertices are \((0,3)\) and \((0,-3)\text{.}\) There are no \(x\)-intercepts.

We construct the central rectangle with dimensions \(2a=4\) and \(2b=6\text{,}\) as shown in the figure. Then we draw the asymptotes through the diagonals of the rectangle. The asymptotes have slopes \(\pm \dfrac{3}{2}\text{.}\) Finally, we sketch the branches of the hyperbola through the vertices and approaching the asymptotes.

We first divide each side by 16 to obtain

The graph is a central hyperbola with \(y\)-intercepts \(2\) and \(-2\text{,}\) as shown in the figure. The slopes of the asymptotes are given by

\begin{gather*} \pm \frac{b}{a}=\pm \frac{2}{4}=\pm \frac{1}{2} \end{gather*}

so the equations of the asymptotes are

\begin{gather*} y= \dfrac{1}{2} x\quad \text{ and } y= \frac{-1}{2} x \end{gather*}

1. The equation \(x^2=6y^2+8\) is equivalent to

   \begin{gather*} x^2-6y^2=8\qquad \text{ or }\qquad \frac{x^2}{\left( \sqrt{8} \right)^2} - \frac{y^2}{\left( \frac{2}{\sqrt{3}} \right)^2}=1 \end{gather*}

   The graph is a hyperbola that opens left and right.

2. The equation \(x^2=\dfrac{4-y^2}{2}\) is equivalent to

   \begin{gather*} 2x^2+y^2=4\qquad \text{ or }\qquad \frac{x^2}{\left( \sqrt{2} \right)^2} + \frac{y^2}{2^2}=1 \end{gather*}

   The graph is an ellipse with major axis on the \(y\)-axis because \(2\gt\sqrt{2} \text{.}\)

1. The graph is a hyperbola with

   \begin{gather*} (h,k) = (3,-2),\quad a = \sqrt{8} = 2\sqrt{2},\quad \text{ and } b=\sqrt{10} \end{gather*}

   Because the \(x^2\)-term is positive, the branches open left and right. The coordinates of the vertices are thus \((3+2\sqrt{2},-2)\) and \((3-2\sqrt{2},-2)\text{,}\) or approximately \((5.8,-2)\) and \((0.2,-2)\text{.}\) The ends of the conjugate axis are \((3,-2+\sqrt{10}\)) and \((3,-2-\sqrt{10}\)), or approximately \((3,1.2)\) and \((3,-5.2)\text{.}\)

   

   The central rectangle is centered at the point \((3,-2)\) and extends to the vertices in the horizontal direction and to the ends of the conjugate axis in the vertical direction. We draw the asymptotes through the opposite corners of the central rectangle, and sketch the hyperbola through the vertices and approaching the asymptotes to obtain the graph shown below.

2. Both asymptotes pass through the center of the hyperbola, \((3,-2)\text{.}\) Their slopes are

   \begin{gather*} \frac{\sqrt{10}}{\sqrt{8}} = \frac{\sqrt{5}}{2}\quad \text{ and } \dfrac{-\sqrt{5}}{2} \end{gather*}

   We substitute these values into the point-slope formula to find the equations

   \begin{gather*} y + 2 = \frac{\sqrt{5}}{2}(x-3)\quad \text{ and } \quad y + 2 = \frac{-\sqrt{5}}{2}(x-3) \end{gather*}

1. We prepare to complete the square by factoring out \(-4\) from the \(x\)-terms.

   \begin{gather*} (y^2+4y\hphantom{-} \underline{ \hphantom{0000}})-4(x^2+2x\hphantom{-}\underline{\hphantom{0000}})=9 \end{gather*}

   We complete the square in \(y\) by adding \(\alert{4}\) to each side of the equation. To complete the square in \(x\) we add 1 to \(x^2+2x\text{,}\) so we add \(-4\cdot \alert{1}\text{,}\) or \(\alert{-4}\) to the right side, to get

   \begin{gather*} (y^2 + 4y + \alert{4}) - 4(x^2 + 2x + \alert{1}) = 9 + \alert{4}~\alert{- 4} \end{gather*}

   or

   \begin{gather*} (y+2)^2 - 4(x+1)^2 = 9 \end{gather*}

   Finally, we divide each side by 9 to obtain the standard form

   \begin{gather*} \frac{(y+2)^2}{9} - \frac{(x+1)^2}{\dfrac{9}{4}} = 1 \end{gather*}

2. The graph is a hyperbola with center at \((-1,-2)\text{.}\) Because the \(y^2\)-term is positive, the transverse axis is parallel to the \(y\)-axis, and

   \begin{gather*} b^2 = 9,\quad a^2 = \frac{9}{4} \end{gather*}

   Thus, \(b=3\) and \(a = \dfrac{3}{2}\text{,}\) and the vertices are \((-1,1)\) and \((-1,5)\text{.}\) The ends of the conjugate axis are \(\left(\dfrac{-5}{2}, -2\right)\) and \(\left(\dfrac{1}{2}, -2\right)\text{.}\)

   

   The central rectangle is centered at \((-1,-2)\text{,}\) as shown in the figure. We draw the asymptotes through the corners of the rectangle, then sketch the hyperbola by starting at the vertices and approaching the asymptotes.

1. The graph is a circle because the coefficients of \(x^2\) and \(y^2\) are equal.

2. The graph is an ellipse because the coefficients of \(x^2\) and \(y^2\) are both positive.

3. The graph is a hyperbola because the coefficients of \(x^2\) and \(y^2\) have opposite signs.

4. The graph is a parabola because \(y\) is of first degree and \(x\) is of second degree.