Question

Graph each equation.

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

Short answer

Part (a)

Part (b)

Part (c)

Part (d)

Part (e)

Part (f)

Part (g)

Part (h)

Part (i)

Part (j)

Step-by-step solution

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,