Chapter 6, Problem 14CR

(a) Find a linear function that contains the points $\left(-2,3\right)$ and $\left(1,-6\right)$. What is the slope? What are the intercepts of the function? Graph the function. Be sure to label the intercepts.

(b) Find a quadratic function with the vertex $\left(1,-6\right)$ that contains the point $\left(-2,3\right)$. What are the intercepts of the function? Graph the function.

(c) Show that there is no exponential function of the form $f\left(x\right)=a{e}^x$ that contains the points $\left(-2,3\right)$ and $\left(1,-6\right)$.

To determine

To calculate: The equation, slope, and intercepts of line containing the points $\left(-2,3\right)$ and $\left(1,-6\right)$. And then, graph it

Answer to Problem 14CR

Solution:

The linear equation passing through points $\left(-2,3\right)$ and $\left(1,-6\right)$ is $y=-3x-3$

The slope of the linear function passing through points $\left(-2,3\right)$ and $\left(1,-6\right)$ is $-3$

The $x-$ intercept is $\left(-1,0\right)$

The $y-$ intercept is $\left(0,-3\right)$

The graph of the linear function is

Explanation of Solution

Given information:

The points $\left(-2,3\right)$ and $\left(1,-6\right)$.

Formula used:

1) Slope formula: The slope between two points $(x_1,y_1)$ and $(x_2,y_2)$, denoted by $m$ is $m=\frac{y_2-y_1}{x_2-x_1}$.

2) Equation of the line using point-slope formula:

The equation of line having slope $m$ and a point $(x_1,y_1)$ is $y-y_1=m\left(x-x_1\right)$

Calculation:

Consider $(x_1,y_1)=(-2,3)$ and $(x_2,y_2)=(1,-6)$

The slope using $(x_1,y_1)=(-2,3)$ and $(x_2,y_2)=(1,-6)$

$m=\frac{-6-3}{1-\left(-2\right)}$

$\Rightarrow m=\frac{-9}{3}$

$\Rightarrow m=-3$

Use the point-slope form the equation of the line is

Use the point $(x_1,y_1)=(-2,3)$ and $m=-3$ in the slope intercept form.

$\Rightarrow y-3=-3\left(x-\left(-2\right)\right)$

$\Rightarrow y-3=-3\left(x+2\right)$

$\Rightarrow y-3=-3x-6$

$\Rightarrow y=-3x-3$

Substitute $y=0$ to find the $x-$ intercept.

$\Rightarrow 0=-3x-3$

$\Rightarrow -3x=3$

$\Rightarrow \frac{-3x}{-3}=\frac{3}{-3}$

$\Rightarrow x=-1$

Thus, the $x-$ intercept is $\left(-1,0\right)$

Substitute $x=0$ to find the $y-$ intercept

$\Rightarrow y=-3\left(0\right)-3$

$\Rightarrow y=-3$

Thus, the $y-$ intercept is $\left(0,-3\right)$

To graph this linear function, follow these steps:

Step 1: Plot the points $\left(-2,3\right)$ and $\left(1,-6\right)$ on the same grid.

Step 2: Sketch a straight line passing through the points $\left(-2,3\right)$ and $\left(1,-6\right)$.

This is the graph of the linear function passing through the points $\left(-2,3\right)$ and $\left(1,-6\right)$ with $x-$ intercept $\left(-1,0\right)$ and $y-$ intercept $\left(0,-3\right)$

To determine

The equation and intercepts of the quadratic function with vertex $\left(1,-6\right)$ and passing through $\left(2,3\right)$, and then graph it

Answer to Problem 14CR

Solution:

The quadratic equation is $y={\left(x-1\right)}^2-6$

The $x-$ intercepts are $x=\sqrt{6}+1$ and $x=-\sqrt{6}+1$

The $y-$ intercept is $\left(0,-5\right)$

The graph of the quadratic function is

Explanation of Solution

Given information:

The vertex is at $\left(1,-6\right)$ and the function is passing through $\left(-2,3\right)$

Explanation:

Vertex form of the quadratic equation with vertex at $\left(h,k\right)$ is $y=a{\left(x-h\right)}^2+k$

Vertex is at $\left(1,-6\right)$. So, $\left(h,k\right)=\left(1,-6\right)$

Substitute $h=1$ and $k=-6$ in the vertex form of the quadratic equation

$\Rightarrow y=a\left(x-1\right)+\left(-6\right)$

$\Rightarrow y=a{\left(x-1\right)}^2-6$

Consider $\left(x,y\right)=\left(-2,3\right)$ and substitute $x=-2$ and $y=3$ in the $y=a{\left(x-1\right)}^2-6$

$\Rightarrow 3=a{\left(-2-1\right)}^2-6$

Solve for $a$

$\Rightarrow 3=a{\left(-3\right)}^2-6$

$\Rightarrow 9=9a$

$\Rightarrow a=1$

From $a=1$, the quadratic equation is $y={\left(x-1\right)}^2-6$

Substitute $y=0$ to find the $x-$ intercept.

$\Rightarrow 0={\left(x-1\right)}^2-6$

Solve for $x$

$\Rightarrow {\left(x-1\right)}^2=6$

$\Rightarrow \left(x-1\right)=\pm \sqrt{6}$

$\Rightarrow x=\sqrt{6}+1\approx 3.45,x=-\sqrt{6}+1\approx -1.45$

Thus, the $x-$ intercepts are $x=\sqrt{6}+1\text{and}x=-\sqrt{6}+1$

Substitute $x=0$ to find the $y-$ intercept

$\Rightarrow y={\left(0-1\right)}^2-6$

$\Rightarrow y=1-6$

$\Rightarrow y=-5$

Thus, the $y-$ intercept is $\left(0,-5\right)$

To graph this quadratic function, follow these steps:

Step 1: Plot the vertex $\left(1,-6\right)$, $x-$ intercepts $x=\sqrt{6}+1\text{and}x=-\sqrt{6}+1$, and $y-$ intercept $\left(0,-5\right)$ on the same grid

Step 2: Plot the point $\left(-2,3\right)$

Step 2: Sketch a smooth curve passing through all points and the shape is like parabola.

This is the graph of the quadratic function with the vertex is at $\left(1,-6\right)$ and the function is passing through $\left(-2,3\right)$

To determine

To show: There is no exponential function of the form $f\left(x\right)=a{e}^x$ that contains the point $\left(-2,3\right)$ and $\left(1,-6\right)$.

Explanation of Solution

Given information:

The points are $\left(-2,3\right)$ and $\left(1,-6\right)$

Formula used:

If $\left(a,b\right)$ is a point on the functio $y=f\left(x\right)$, then $b=f\left(a\right)$

Proof:

Replace $f\left(x\right)$ with $y$ in the exponential form $f\left(x\right)=a{e}^x$

$\Rightarrow y=a{e}^x$

Using, “If $\left(a,b\right)$ is a point on the function $y=f\left(x\right)$, then $b=f\left(a\right)$”

Consider $\left(x,y\right)=\left(-2,3\right)$ and substitute $x=-2$ and $y=3$ in $y=a{e}^x$

$\Rightarrow 3=a{e}^{-2}$

Solve for $a$

$\Rightarrow a=\frac{3}{e^{-2}}=3e^2$

Consider $\left(x,y\right)=\left(1,-6\right)$ and substitute $x=1$ and $y=-6$ in $y=a{e}^x$

$\Rightarrow -6=a{e}^1$

Solve for $a$

$\Rightarrow a=-\frac{6}{e}$

For point $\left(-2,3\right)$, the value of $a$ is $3e^2$ and for the point $\left(1,-6\right)$, the value of $a$ is $-\frac{6}{e}$. Because of the two different values of $a$, there is no exponential function containing the point $\left(-2,3\right)$ and $\left(1,-6\right)$.

Hence, proved