Q 11. - Precalculus Enhanced with Graphing Utilities

Question

Graph each equation.

$\begin{array}{l}\left(a\right)y=3x+6 \left(b\right)x^2+y^2=4 \\ \left(c\right)y=x^3 \left(d\right)y=\frac{1}{x}\\ \left(e\right)y=\sqrt{x} \left(f\right)y=e^x\\ \left(g\right)y=\ln x \left(h\right)2x^2+5y^2=1\\ \left(i\right)x^2-3y^2=1 \left(j\right)x^2-2x-4y+1=0\end{array}$

Step-by-Step Solution

Answer

Part (a)

Part (b)

Part (c)

Part (d)

Part (e)

Part (f)

Part (g)

Part (h)

Part (i)

Part (j)

Part (a) Step 1. Given information.

Consider the given information.

$\begin{array}{l}\left(a\right)y=3x+6 \left(b\right)x^2+y^2=4 \\ \left(c\right)y=x^3 \left(d\right)y=\frac{1}{x}\\ \left(e\right)y=\sqrt{x} \left(f\right)y=e^x\\ \left(g\right)y=\ln x \left(h\right)2x^2+5y^2=1\\ \left(i\right)x^2-3y^2=1 \left(j\right)x^2-2x-4y+1=0\end{array}$

Part (a) Step 2. Draw the graph.

Consider the given function.

$y=3x+6$

Find the intercepts.

$\begin{array}{rcl}y& =& 3\left(0\right)+6\\ & =& 6\end{array}$

And,

$\begin{array}{rcl}0& =& 3x+6\\ x& =& -2\end{array}$

The x and y-intercepts are $\left(-2,0\right),\left(0,6\right)$. Now, draw the graph.

Part (b) Step 1. Draw the graph.

Consider the given function.

$x^2+y^2=4$

Find the intercepts.

$\begin{array}{rcl}{x}^2+0^2& =& 4\\ x& =& -2,2\end{array}$

And,

$\begin{array}{rcl}{0}^2+y^2& =& 4\\ y& =& -2,2\end{array}$

The x and y-intercepts are $\left(-2,0\right),\left(2,0\right) and \left(0,2\right),\left(0,-2\right)$. Now, draw the graph.

Part (c) Step 1. Draw the graph.

Consider the given function.

$y=x^3$

Find the intercepts.

$\begin{array}{rcl}y& =& 0^3\\ & =& 0\end{array}$

And,

$\begin{array}{rcl}0& =& x^3\\ x& =& 0\end{array}$

The x and y-intercepts are  $\left(0,0\right) and \left(0,0\right)$. Now, draw the graph,

Part (d) Step 1. Draw the graph.

Consider the given function.

$y=\frac{1}{x}$

The intercept for the function, does not exist. Use the graphing utility to draw the graph.

Part (e) Step 1. Draw the graph.

Consider the given function.

$y=\sqrt{x}$

Find the intercepts.

$\begin{array}{rcl}y& =& \sqrt{0}\\ & =& 0\end{array}$

And,

$\begin{array}{rcl}0& =& \sqrt{x}\\ x& =& 0\end{array}$

The x and y-intercepts are $\left(0,0\right) and \left(0,0\right)$. Now, draw the graph by using graphing utility,

Part (f) Step 1. Draw the graph.

Consider the given function.

$y=e^x$

Find the intercepts.

$\begin{array}{rcl}y& =& e^0\\ & =& 1\end{array}$

And, the x -axis does not exist. Thus the y-intercept is $\left(0,1\right)$. Use the graphing utility to draw the graph.

Part (g) Step 1. Draw the graph.

Consider the given function.

$y=\ln x$

Find the intercepts.

$\begin{array}{rcl}\ln x& =& 0\\ x& =& e^0\\ & =& 1\end{array}$

And, the y-intercept does not exist. Use the graphing utility to draw the graph.

Part (h) Step 1. Draw the graph.

Consider the given function.

$2x^2+5y^2=1$

Find the intercepts.

$\begin{array}{rcl}2x^2+5{\left(0\right)}^2& =& 1\\ 2x^2& =& 1\\ x& =& \sqrt{\frac{{\displaystyle 1}}{{\displaystyle 2}}},\hspace{0.17em}-\sqrt{\frac{{\displaystyle 1}}{{\displaystyle 2}}}\end{array}$

And,

$\begin{array}{rcl}2{\left(0\right)}^2\hspace{0.17em}+\hspace{0.17em}5y^2\hspace{0.17em}& =& \hspace{0.17em}1\\ \hspace{0.17em}5y^2\hspace{0.17em}& =& \hspace{0.17em}1\\ y& =& \sqrt{\frac{1}{5}},\hspace{0.17em}-\sqrt{\frac{1}{5}}\end{array}$

The x and y-intercepts are  $\left(\sqrt{\frac{1}{2}},0\right),\left(-\sqrt{\frac{1}{2}},0\right) \mathrm{and} \left(0,\sqrt{\frac{1}{5}}\right),\left(0,-\sqrt{\frac{1}{5}}\right)$. Now draw the graph.

Part (i) Step 1. Draw the graph.

Consider the given function.

$x^2-3y^2=1$

Find the intercepts.

$\begin{array}{rcl}{x}^2-3{\left(0\right)}^2& =& 1\\ x^2& =& 1\\ x& =& -1,1\end{array}$

And, the y-intercept does not exist. Now, draw the graph by using a graphing utility.

Part (j) Step 1. Draw the graph.

Consider the given function.

$\begin{array}{rcl}{x}^2-2x-4y+1& =& 0\\ {\left(x-1\right)}^2& =& 4y\end{array}$

Find the intercepts.

$\begin{array}{l}{\left(x-1\right)}^2=4\left(0\right)\\ x-1=0\\ x=1\end{array}$

And,

$\begin{array}{l}{\left(0-1\right)}^2=4y\\ 1=4y\\ y=\frac{1}{4}\end{array}$

The x and y-intercepts are  $\left(1,0\right) and \left(0,\frac{1}{4}\right)$. draw the graph by using the graphing utility,