Chapter 3.1, Problem 13E

a.

To determine

To Express: The quadratic function in standard form.

Answer to Problem 13E

the quadratic function is expressed in standard form as $f\left(x\right)=1{\left(x+2\right)}^2-1$

Explanation of Solution

Given: The function is $f\left(x\right)=\left(x^2+4x+3\right)$

Calculation:

The quadratic function $f\left(x\right)=\left(x^2+4x+3\right)$ is expressed in standard form as:

  $f\left(x\right)=a{\left(x-h\right)}^2+k$ , by completing the square. The graph of the function f is a parabola with vertex $\left(h,k\right)$

The parabola opens upwards if $a>0$ .

Solve the function:

  $\begin{array}{l}f\left(x\right)=\left(x^2+4x+3\right)\\ f\left(x\right)=1\left(x^2+4x\right)+3\text{ [factor 1 from the }x\text{ terms]}\\ f\left(x\right)=1\left(x^2+4x+4\right)+3-\text{1}\cdot 4\text{ [complete the square : add 4 inside parentheses, subtract 1}\cdot 4\text{ outside]}\\ f\left(x\right)=1{\left(x+2\right)}^2-1\text{ [factor and multiply]}\end{array}$

On comparing the above equation with standard form $f\left(x\right)=a{\left(x-h\right)}^2+k$ ,

Therefore, the quadratic function is expressed in standard form as $f\left(x\right)=1{\left(x+2\right)}^2-1$

b.

To determine

To Find: The vertex, $x-\text{intercept}$ and $y-\text{intercept}$

Answer to Problem 13E

the vertex is $\left(h,k\right)=\left(-2,-1\right)$ .

  $x\text{-intercept}=-1\text{ and }-3$

  $y\text{-intercept}=f\left(0\right)=3$

Explanation of Solution

Given: The function is $f\left(x\right)=\left(x^2+4x+3\right)$

Calculation:

The quadratic function $f\left(x\right)=\left(x^2+4x+3\right)$ is expressed in standard form as:

  $f\left(x\right)=1{\left(x+2\right)}^2-1$

by completing the square. The graph of the function f is a parabola with vertex $\left(h,k\right)$ , the vertex is $\left(h,k\right)=\left(-2,-1\right)$ .

The $y\text{-intercept}$ is given when $x=0$ , so $y\text{-intercept}=f\left(0\right)=3$

The $x\text{-intercept}$ is given when $y=0$ , so $x^2+4x+3=0$ which implies $x\left(x+4\right)+3=0$

Therefore, $x\text{-intercept}=-1\text{ and }-3$

c.

To determine

To Sketch: The graph of the quadratic function.

Explanation of Solution

Given: The function is $f\left(x\right)=\left(x^2+4x+3\right)$

Graph:

The standard form of the function is:

  $f\left(x\right)=1{\left(x+2\right)}^2-1$

From the standard form it is observed that the graph is a parabola that opens upward and has vertex $\left(-2,-1\right)$ . As an aid to sketching the graph, find the intercepts. The $y\text{-intercept}=f\left(0\right)=3$ and the $x\text{-intercept}=-1\text{ and }-3$ . The graph f is sketched in the figure below.

Use graphing calculator to graph the function: $f\left(x\right)=\left(x^2+4x+3\right)$