Chapter 3, Problem 23RE

(a).

To determine

To graph: Quadratic function with its vertex, axis of symmetry, $y-$ intercept or $x-$ intercept.

Answer to Problem 23RE

Function using the vertex and the $x-$ intercept graph is:

  

Explanation of Solution

Given:

  $f\left(x\right)=3x^2+4x-1$

Calculation:

The quadratic function $3x^2+4x-1$ is form $f\left(x\right)=a{x}^2+bx+c$ .

So, $a=3,b=4,c=-1$

The graph of a quadratic function $f\left(x\right)=a{x}^2+bx+c$ opens up If $a>0$ and opens down if $a<0$ .

Here, function has a $a=3$ which is greater than $0$ .

So the graph opens up.

To find the vertex,

The coordinates of the vertex of a quadratic function of the form $f\left(x\right)=a{x}^2+bx+c$ is $\left(\frac{-b}{2a},f\left(\frac{-b}{2a}\right)\right)$ .

Evaluating $\frac{-b}{2a}$ to find the $x-$ coordinate of the vertex. Substitute the values $a=3,b=4$ in $\frac{-b}{2a}=\frac{-4}{2\left(3\right)}=-\frac{2}{3}$

Evaluating $f$ at $\left(\frac{-b}{2a}\right)$ to find y-coordinate of the vertex. So put $-\frac{2}{3}$ for $\left(\frac{-b}{2a}\right)$ and evaluate the function.

  $\begin{array}{l}f\left(\frac{-b}{2a}\right)=f\left(\frac{1}{2}\right)\\ =3\left(\frac{4}{9}\right)+4\left(-\frac{2}{3}\right)-1\\ =-\frac{7}{3}\end{array}$

Thus, the vertex is at $\left(-\frac{2}{3},-\frac{7}{3}\right)$ .

The axis of symmetry is the line where

  $x=\frac{-b}{2a}$

Thus, $x=-\frac{2}{3}$ is the axis of symmetry.

To find $y-$ intercept,

  $y-$ Intercept is the value of $f$ at $x=0$

Here, $f\left(0\right)=0+0-1=-1$

Therefore, the graph has the $y-$ intercept at $-1$ .

Find the discriminant $b^2-4ac$ of the equation by substituting $a=3,b=4,c=-1$ .

  $\begin{array}{l}{b}^2-4ac={\left(4\right)}^2-4\left(3\right)\left(-1\right)\\ =16+12\\ =28\end{array}$

In case discriminant $b^2-4ac>0$ , the graph of $f\left(x\right)=a{x}^2+bx+c$ has two distinct $x-$ intercepts.

Therefore, the graph of the function crosses the $x-$ axis in two places.

To find out the $x-$ intercept,

The $x-$ intercept can be obtained by solving the quadratic equation of the from $a{x}^2+bx+c=0$

  $3x^2+4x-1=0$

Substitute, $a=3,b=4,c=-1$ in the quadratic formula and solve the equation.

  $\begin{array}{l}\frac{-b\pm \sqrt{b^2-4ac}}{2a}=\frac{-4\pm \sqrt{28}}{2\left(3\right)}\\ =\frac{-4\pm 2\sqrt{7}}{6}\\ =0.215or-1.548\\ =0.22or-1.55\end{array}$

Function using the vertex and the $x-$ intercept graph is:

  

Conclusion:

Quadratic function with its vertex is $\left(-\frac{2}{3},-\frac{7}{3}\right)$ , axis of symmetry, $y-$ intercept or $x-$ intercept.

(b).

To determine

Domain and range of the function.

Answer to Problem 23RE

Domain is the interval $\left(-\infty ,\infty \right)$ and range of $f$ is the interval $\left[-\frac{7}{3},\infty \right)$ .

Explanation of Solution

Given:

  $f\left(x\right)=3x^2+4x-1$

Calculation:

Find the domain and range of the function, the domain of $f$ is the set of all real numbers.

From the graph, find the domain is the interval

  $\left(-\infty ,\infty \right)$ and range of $f$ is the interval $\left[-\frac{7}{3},\infty \right)$ .

Conclusion:

Thus, domain is the interval $\left(-\infty ,\infty \right)$ and range of $f$ is the interval $\left[-\frac{7}{3},\infty \right)$ .

(c).

To determine

The function is increasing and where it is decreasing.

Answer to Problem 23RE

The function $f$ is decreasing on the interval $\left(-\infty ,-\frac{2}{3}\right)$ and increasing on the interval $\left(-\frac{2}{3},\infty \right)$

Explanation of Solution

Given:

  $f\left(x\right)=3x^2+4x-1$

Calculation:

Interval where the function $f$ is decreasing.

Also, find where the function is increasing.

From the graph, the function $f$ is decreasing on the interval $\left(-\infty ,-\frac{2}{3}\right)$ and increasing on the interval $\left(-\frac{2}{3},\infty \right)$ .

Conclusion:

Thus, the function $f$ is decreasing on the interval $\left(-\infty ,-\frac{2}{3}\right)$ and increasing on the interval $\left(-\frac{2}{3},\infty \right)$ .