Chapter 10, Problem 6CR

(a)

To determine

The equation for the following graph:

Answer to Problem 6CR

Solution:

The equation of the line given in graph is $y=2x-2$.

Explanation of Solution

Given information:

The graph

Explanation:

From the above graph, it is seen that the line passes through the points $\left(0,-2\right)$ and $\left(1,0\right)$.

The general equation of a straight line passing through two points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ is

$y-y_1=m\left(x-x_1\right)$, where $m$ is the slope and $m=\frac{y_2-y_1}{x_2-x_1}$.

Here, the points are $\left(x_1,y_1\right)=\left(0,-2\right)$ and $\left(x_2,y_2\right)=\left(1,0\right)$

Slope of the line is $m=\frac{0-\left(-2\right)}{1-0}=2$

Then the equation of the line passing through the points $\left(0,-2\right)$ and $\left(1,0\right)$ is

$y-\left(-2\right)=2\left(x-0\right)$

$\Rightarrow y+2=2x$

$\Rightarrow y=2x-2$

Hence the equation of the line given in graph is $y=2x-2$.

(b)

To determine

The equation for the following graph of circle:

Answer to Problem 6CR

Solution:

The equation of the circle given in the graph is $x^2+y^2-4x=0$

Explanation of Solution

Given information:

The graph

Explanation:

From the above graph, it is seen that the center of the given circle is $\left(2,0\right)$ and radius is $2$.

The general equation of the circle having center at $\left(h,k\right)$ and radius $r$ is

${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$.

Here, $\left(h,k\right)=\left(2,0\right)$ and $r=2$.

Then, the equation of the given circle by substituting the above values is

${\left(x-2\right)}^2+{\left(y-0\right)}^2=2^2$

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

$\Rightarrow x^2+4-4x+y^2=4$

$\Rightarrow x^2+y^2-4x=0$

Hence, the equation of the circle given in graph is $x^2+y^2-4x=0$.

(c)

To determine

The equation for the following graph of ellipse:

Answer to Problem 6CR

Solution:

The equation of the ellipse given in the graph is $\frac{x^2}{9}+\frac{y^2}{4}=1$

Explanation of Solution

Given information:

The graph

Explanation:

From the above graph, it is seen that the center of the given ellipse is $\left(0,0\right)$, vertices are at $\left(\pm 3,0\right)$ and $y-$ intercepts are at $\left(0,\pm 2\right)$.

The general equation of the ellipse having center at $\left(h,k\right)$ and vertices at $\left(\pm a,0\right)$ and $y-$ intercepts are at $\left(0,\pm b\right)$ is $\frac{{\left(x-h\right)}^2}{a^2}+\frac{{\left(y-k\right)}^2}{b^2}=1$.

Here, $\left(h,k\right)=\left(0,0\right),\left(\pm a,0\right)=\left(\pm 3,0\right)$ and $\left(0,\pm b\right)=\left(0,\pm 2\right)$.

Then, the equation of the given ellipse by substituting the above values is

$\frac{{\left(x-0\right)}^2}{3^2}+\frac{{\left(y-0\right)}^2}{2^2}=1$

$\Rightarrow \frac{x^2}{9}+\frac{y^2}{4}=1$

Hence, the equation of the ellipse given in the graph is $\frac{x^2}{9}+\frac{y^2}{4}=1$.

(d)

To determine

The equation for the following graph of parabola:

Answer to Problem 6CR

Solution:

The equation of the parabola given in graph is $y=2{\left(x-1\right)}^2$.

Explanation of Solution

Given information:

The graph

Explanation:

From the above graph, it is seen that the center of the given parabola is $\left(1,0\right)$ and passes through the point $\left(0,2\right)$ and the parabola opens up.

The general equation of the parabola which opens up having center at $\left(h,k\right)$ is ${\left(x-h\right)}^2=4a\left(y-k\right)$.

Here, $\left(h,k\right)=\left(1,0\right)$

Then, the equation of the given parabola by substituting the above values is

${\left(x-1\right)}^2=4a\left(y-0\right)$

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

As the parabola passes through the point $\left(0,2\right)$, this point satisfies the equation.

Substitute this point $\left(0,2\right)$ in the equation ${\left(x-1\right)}^2=4ay$

$\Rightarrow {\left(0-1\right)}^2=4a\left(2\right)$

$\Rightarrow 1=8a$

$\Rightarrow a=\frac{1}{8}$

Now, to get the equation of the parabola, substitute the value of $a$ in ${\left(x-1\right)}^2=4ay$

$\Rightarrow {\left(x-1\right)}^2=4\times \frac{1}{8}\times y$

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

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

Hence, the equation of the parabola given in the graph is $y=2{\left(x-1\right)}^2$

(e)

To determine

The equation for the following graph of hyperbola:

Answer to Problem 6CR

Solution:

The equation of the hyperbola given in the graph is $y^2-\frac{x^2}{9}=1$.

Explanation of Solution

Given information:

The graph

Explanation:

From the above graph, it is seen that the center of the given hyperbola is $\left(0,0\right)$, vertices at

$\left(0,\pm 1\right)$, passes through the point $\left(3,2\right)$ and the transverse axis is along the $y-$ axis.

The general equation of the hyperbola whose transverse axis is along the $y-$ axis having center at $\left(h,k\right)$, vertices at $\left(0,\pm a\right)$ is $\frac{{\left(y-k\right)}^2}{a^2}-\frac{{\left(x-h\right)}^2}{b^2}=1$

Here, $\left(h,k\right)=\left(0,0\right),\left(0,\pm a\right)=\left(0,\pm 1\right)$

Then, the equation of the given hyperbola by substituting the above values is

$\frac{{\left(y-0\right)}^2}{{\left(1\right)}^2}-\frac{{\left(x-0\right)}^2}{b^2}=1$

$\Rightarrow y^2-\frac{x^2}{b^2}=1$

As the hyperbola passes through the point $\left(3,2\right)$, this point satisfies the equation.

Substitute this point $\left(3,2\right)$ in the equation $y^2-\frac{x^2}{b^2}=1$ to find $b^2$

$2^2-\frac{3^2}{b^2}=1$

$\Rightarrow 4-\frac{9}{b^2}=1$

$\Rightarrow \frac{9}{b^2}=4-1$

$\Rightarrow b^2=\frac{9}{3}$

$\Rightarrow b^2=3$

Now, to get the equation of the parabola, substitute the value of $b^2$ in $y^2-\frac{x^2}{b^2}=1$

$\Rightarrow y^2-\frac{x^2}{9}=1$

Hence, the equation of the hyperbola given in the graph is $y^2-\frac{x^2}{9}=1$.

(f)

To determine

The equation for the following graph of exponential function:

Answer to Problem 6CR

Solution:

The equation of the exponential function given in graph is $y=4^x$

Explanation of Solution

Given information:

The graph

Explanation:

From the above graph, it is seen that the exponential function has $y-$ intercept at $\left(0,1\right)$ and passes through the point $\left(1,4\right)$

The general equation of the exponential function is $y=A{b}^x$

Plug the $y-$ intercept $\left(0,1\right)$ in $y=A{b}^x$

$\Rightarrow 1=A{b}^0$

$\Rightarrow 1=A$

Then, the equation of the exponential function becomes $y=b^x$

The equation of the given exponential function passes through the point $\left(1,4\right)$

Substitute this point $\left(1,4\right)$ in the equation $y=b^x$ to find $b$.

$\Rightarrow 4=b^1$

$\Rightarrow b=4$

Now, to get the equation of the exponential function, substitute the value of $b$ in $y=b^x$

$\Rightarrow y=4^x$

Hence, the equation of the exponential function given in the graph is $y=4^x$