Chapter 11, Problem 11CR

(a)

To determine

To graph: The equation $y=3x+6$.

Explanation of Solution

Given information:

The equation $y=3x+6$.

Graph:

The equation $y=3x+6$ is a line.

To graph the line $y=3x+6$, find $x-$ intercept and $y-$ intercept a line.

The $x-$ intercept occurs when $y=0$.

$\Rightarrow 3x+6=0$

By solving this equation $x=-2$.

Hence $x-$ intercept of the line is $\left(-2,0\right)$.

The $y-$ intercept occur where $x=0$.

$\Rightarrow 3\left(0\right)+6=y$

By solving this equation, $y=6$.

Hence $y-$ intercept of the line is $\left(0,6\right)$.

Sketch a line passing through the points $\left(-2,0\right)$ and $\left(0,6\right)$, get the graph of a line $y=3x+6$ as

Interpretation:

The graph shows the graph of the line $y=3x+6$. The line $y=3x+6$ has intercepts $\left(-2,0\right)$ and $\left(0,6\right)$.

(b)

To determine

To graph: The equation $x^2+y^2=4$.

Explanation of Solution

Given information:

The equation $x^2+y^2=4$.

Graph:

The equation $x^2+y^2=4$ is a circle with radius $2$ and center is origin $\left(0,0\right)$.

Hence, the points on the circle are $\left(2,0\right),\left(-2,0\right),\left(0,2\right)$ and $\left(0,-2\right)$.

The graph of a circle $x^2+y^2=4$ is,

Interpretation:

The graph shows the graph of a circle $x^2+y^2=4$. The circle has a center at origin and the radius is $2$.

(c)

To determine

To graph: The equation $y=x^3$.

Explanation of Solution

Given information:

The equation $y=x^3$.

Graph:

To graph the function $y=x^3$, find the points $\left(x,x^3\right)$ on the graph of $y=x^3$, sketch the curve passing through these points.

For each point $\left(x,x^3\right)$ on the graph of $y=x^3$, indicated in the below table.

$\begin{array}{ccc}x& y=x^3& \text{points}\\ 0& 0& \left(0,0\right)\\ 1& 1& \left(1,1\right)\\ 2& 8& \left(2,8\right)\\ 3& 27& \left(3,27\right)\end{array}$

The graph of the function $y=x^3$ is,

Interpretation:

The graph shows the graph of anequation $y=x^3$. The curve $y=x^3$ passing through the origin $\left(0,0\right)$.

(d)

To determine

To graph: The equation $y=\frac{1}{x}$.

Explanation of Solution

Given information:

The equation is $y=\frac{1}{x}$.

Graph:

To graph the function $y=\frac{1}{x}$, find the points $\left(x,\frac{1}{x}\right)$ on the graph of $y=\frac{1}{x}$ then sketch the curve passing through these points.

For each point $\left(x,\frac{1}{x}\right)$ on the graph of $y=\frac{1}{x}$, indicated in the below table.

$\begin{array}{ccc}x& y=\frac{1}{x}& \left(x,y\right)\\ 1& 1& \left(1,1\right)\\ -1& -1& \left(-1,-1\right)\\ 2& 0.5& \left(2,0.5\right)\\ -2& -0.5& \left(-2,-0.5\right)\end{array}$

The graph of the function $y=\frac{1}{x}$ is,

Interpretation:

The graph shows the graph of an equation $y=\frac{1}{x}$. By looking, the graph it is observed that the curve $y=\frac{1}{x}$ is symmetric with respect to origin.

(e)

To determine

To graph: The equation $y=\sqrt{x}$.

Explanation of Solution

Given information:

The equation is $y=\sqrt{x}$.

Graph:

To graph the function $y=\sqrt{x}$, find the points $\left(x,\sqrt{x}\right)$ on the graph of $y=\sqrt{x}$ sketch the curve passing through these points.

For each point $\left(x,\sqrt{x}\right)$ on the graph of $y=\sqrt{x}$, indicated in the below table.

$\begin{array}{ccc}x& y=\sqrt{x}& \left(x,y\right)\\ 1& 1& \left(1,1\right)\\ 4& 2& \left(4,2\right)\\ 9& 3& \left(9,3\right)\\ 16& 4& \left(16,4\right)\end{array}$

The graph of the function $y=\sqrt{x}$ is,

Interpretation:

The graph shows the graph of an equation $y=\sqrt{x}$. The function is defined for only positive real numbers.

(f)

To determine

To graph: The equation $y=e^x$.

Explanation of Solution

Given information:

The equation is $y=e^x$.

Graph:

To graph the function $y=e^x$, find the points $\left(x,e^x\right)$ on the graph of $y=e^x$ sketch the curve passing through these points.

For each point $\left(x,e^x\right)$ on the graph of $y=e^x$, indicated in the below table.

$\begin{array}{ccc}x& y=e^x& \left(x,y\right)\\ 0& 1& \left(0,1\right)\\ 1& e& \left(1,e\right)\\ 2& 7.389& \left(2,7.389\right)\\ -1& 0.368& \left(-1,0.368\right)\end{array}$

The graph of the function $y=e^x$ is,

Interpretation:

The graph shows the graph of an equation $y=e^x$. It is observed that as $x$ value increases $y$ also increases.

(g)

To determine

To graph: The equation $y=\ln x$.

Explanation of Solution

Given information:

The equation is $y=\ln x$.

Graph:

To graph the function $y=\ln x$, find the points $\left(x,\ln x\right)$ on the graph of $y=\ln x$ sketch the curve passing through these points.

For each point $\left(x,\ln x\right)$ on the graph of $y=\ln x$, indicated in the below table.

$\begin{array}{ccc}x& y=\ln x& \left(x,y\right)\\ 1& 0& \left(1,0\right)\\ e& 1& \left(e,1\right)\\ 2& 0.693& \left(2,0.693\right)\\ 4& 1.386& \left(4,1.386\right)\end{array}$

The graph of the function $y=\ln x$ is,

Interpretation:

The graph shows the graph of an equation $y=\ln x$. It is observed that as $x$ value increases $y$ also increases.

(h)

To determine

To graph: The equation $2x^2+5y^2=1$.

Explanation of Solution

Given information:

The equation is $2x^2+5y^2=1$.

Graph: Let, $2x^2+5y^2=1$.

An equation $2x^2+5y^2=1$, can be written as $\frac{x^2}{\frac{1}{2}}+\frac{y^2}{\frac{1}{5}}=1$ that is ${\left(\frac{x}{\frac{\sqrt{2}}{2}}\right)}^2+{\left(\frac{y}{\frac{\sqrt{5}}{5}}\right)}^2=1$.

The equation represent hyperbola with center origin.

The points on the hyperbola $2x^2+5y^2=1$ indicated in the below table.

$\begin{array}{ccc}x& y& \left(x,y\right)\\ \frac{\sqrt{2}}{2}& 0& \left(\frac{\sqrt{2}}{2},0\right)\\ -\frac{\sqrt{2}}{2}& 0& \left(-\frac{\sqrt{2}}{2},0\right)\\ 0& \frac{\sqrt{5}}{5}& \left(0,\frac{\sqrt{5}}{5}\right)\\ 0& -\frac{\sqrt{5}}{5}& \left(0,-\frac{\sqrt{5}}{5}\right)\end{array}$

The graph of the function $2x^2+5y^2=1$ is,

Interpretation:

The graph shows the graph of an equation $2x^2+5y^2=1$. The graph is symmetric with respect to $x-$ axis, $y-$ axis and origin.

(i)

To determine

To graph: The equation $x^2-3y^2=1$.

Explanation of Solution

Given information:

The equation is $x^2-3y^2=1$.

Graph: Let $x^2-3y^2=1$

An equation $x^2-3y^2=1$, can be written as $\frac{x^2}{1}-\frac{y^2}{\frac{1}{3}}=1$ that is ${\left(\frac{x}{1}\right)}^2-{\left(\frac{y}{\frac{\sqrt{3}}{3}}\right)}^2=1$.

The equation represent hyperbola with center origin.

The points on the hyperbola $x^2-3y^2=1$ indicated in the below table.

$\begin{array}{ccc}x& y& \left(x,y\right)\\ 1& 0& \left(1,0\right)\\ -1& 0& \left(-1,0\right)\\ 2& 1& \left(2,1\right)\\ 2& -1& \left(2,-1\right)\\ -2& 1& \left(-2,1\right)\\ -2& -1& \left(-2,-1\right)\end{array}$

The graph of the function $x^2-3y^2=1$ is,

Interpretation:

The graph shows the graph of an equation $x^2-3y^2=1$. The graph is symmetric with respect to $x-$ axis, $y-$ axis and origin.

(j)

To determine

To graph: The equation $x^2-2x-4y+1=0$.

Explanation of Solution

Given information:

The equation is $x^2-2x-4y+1=0$.

Graph: Let $x^2-2x-4y+1=0$

By combining the terms,

$\left(x^2-2x+1\right)-4y=0$

Adding $4y$ from both sides,

$\Rightarrow \left(x^2-2x+1\right)=4y$

$\Rightarrow {\left(x-1\right)}^2=4y$

$\Rightarrow y=\frac{1}{4}{\left(x-1\right)}^2$

An equation, $y=\frac{1}{4}{\left(x-1\right)}^2$ represent a upward parabola.

The points on the parabola $y=\frac{1}{4}{\left(x-1\right)}^2$ indicated in the below table.

$\begin{array}{ccc}x& y& \left(x,y\right)\\ 1& 0& \left(1,0\right)\\ -1& 1& \left(-1,0\right)\\ 0& \frac{1}{4}& \left(0,\frac{1}{4}\right)\\ 2& \frac{1}{4}& \left(2,\frac{1}{4}\right)\\ 3& 1& \left(3,1\right)\end{array}$

The graph of the function $x^2-2x-4y+1=0$ is,

Interpretation:

The graph shows the graph of an equation $x^2-2x-4y+1=0$. The equation represent upward parabola.