Chapter 3, Problem 19RE

(a).

To determine

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

Answer to Problem 19RE

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

  

Explanation of Solution

Given:

  $f\left(x\right)=-4x^2+4x$

Calculation:

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

So, $a=-4,b=4$

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 $a=-4$ which is less than $0$ .

So the graph opens down.

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 $a=-4,b=4$ in $\frac{-b}{2a}=\frac{\left(-4\right)}{2\left(4\right)}=\frac{1}{2}$

Evaluating $f$ at $\left(\frac{-b}{2a}\right)$ to find y-coordinate of the vertex. So put $\frac{1}{2}$ for $\left(\frac{-b}{2a}\right)$ and evaluate the function.

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

Thus, the vertex is at $\left(\frac{1}{2},1\right)$ .

The axis of symmetry is the line where $x=\frac{-b}{2a}$

Thus, $x=\frac{1}{2}$ is the axis of symmetry.

To find $y-$ intercept,

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

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

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

Find the discriminant $b^2-4ac$ of the equation by substituting $a=-4,b=4,c=0$ .

  $\begin{array}{l}{b}^2-4ac={\left(4\right)}^2-4\left(-4\right)\left(0\right)\\ =16\end{array}$

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 $a{x}^2+bx+c=0$

  $-4x^2+4x=0$

Factor out $-4x$ from $-4x^2+4x=0$ .

  $-4x\left(x-1\right)=0$

Apply the zero product principle,

Thus, either $-4x=0$ or $x-1=0$ both.

Consider $-4x=0$

  $\begin{array}{l}\frac{-4x}{-4}=0\\ x=0\end{array}$

Consider, $x-1=0$

Add $1$ to both sides of the equation.

  $\begin{array}{l}x+1-1=0+1\\ x=1\end{array}$

Thus, the $x-$ intercepts are $0$ and $1$ .

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

  

Conclusion:

Quadratic function with its vertex is $\left(1/2,1\right)$ , axis of symmetry, $y-$ intercept or $x-$ intercept.

(b).

To determine

Domain and range of the function.

Answer to Problem 19RE

Domain is the interval $\left(-\infty ,\infty \right)$ and range of $f$ is the interval $\left(-\infty ,1\right]$ .

Explanation of Solution

Given:

  $f\left(x\right)=-4x^2+4x$

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 $\left(-\infty ,1\right]$ .

Conclusion:

Thus, domain is the interval $\left(-\infty ,\infty \right)$ and range of $f$ is the interval $\left(-\infty ,1\right]$ .

(c).

To determine

The function is increasing and where it is decreasing.

Answer to Problem 19RE

The function $f$ is decreasing on the interval $\left(\infty ,\frac{1}{2}\right)$ and increasing on the interval $\left(\frac{1}{2},\infty \right)$ .

Explanation of Solution

Given:

  $f\left(x\right)=-4x^2+4x$

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 $\left(\infty ,\frac{1}{2}\right)$ and increasing on the interval $\left(\frac{1}{2},\infty \right)$ .

Conclusion:

Thus, the function $f$ is decreasing on the interval $\left(\infty ,\frac{1}{2}\right)$ and increasing on the interval $\left(\frac{1}{2},\infty \right)$ .