Q. 37 - Precalculus Enhanced with Graphing Utilities

Question

(a).  Graph each quadratic function by determining whether its graph opens up or down and by finding its vertex, axis of symmetry, y-intercept, and x-intercepts, if any. 

(b).  Determine the domain and the range of the function. 

(c).  Determine where the function is increasing and where it is decreasing. 

Verify your results using a graphing utility.

Function -  $F\left(x\right)=x^2+2x+1$

Step-by-Step Solution

Answer

(a). 

• The graph is open downwards.
• The vertex of graph is - (-1, 0)
• The axis of symmetry passes through the vertex of the function and is parallel to the y-axis.

(b). 

• The domain is -  $\mathrm{ℝ}$
• The range is  -  $y\in [0, +\infty )$

(c).

• Function increasing for -  $x\in [-1, +\infty )$
• Function decreasing for -  $x\in (-\infty , -1]$

Part (a). Step 1. Given information

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

Part (a). Step 2. To Find

Graph each quadratic function by determining whether its graph opens up or down and by finding its vertex, axis of symmetry, y-intercept, and x-intercepts, if any.

Part (a). Step 3. Explanation

Remind yourself that the quadratic function is given by -   $f\left(x\right)=a{x}^2+bx+c, a\ne 0$

Since $\mathit{a}\mathbf{<}\mathbf{0}$,We can conclude that the function open downwards. we are looking the vertex as $[-\frac{b}{2a},f(-\frac{b}{2a}\left)\right]$, Lets us find the vertex.

$$\begin{gathered} -\frac{b}{2a}=-\frac{2}{2\times 1} \\ -\frac{b}{2a}=-1 \end{gathered}$$

$$\begin{gathered} f(-\frac{b}{2a})=f(-1) \\ f(-\frac{b}{2a})={(-1)}^2+2\times (-1)+1 \\ f(-\frac{b}{2a})=1-2+1 \\ f(-\frac{b}{2a})=0 \end{gathered}$$

Vertex is -   (-1, 0)

Determine the Axis of symmetric -

$$\begin{gathered} x=-\frac{b}{2a} \\ x=-\frac{2}{2\times 1} \\ x=-1 \end{gathered}$$

The axis of symmetric passes through the vertex of the function and is parallel to the y-axis.

Determine the point at which the function intersects the x-axis -

$$\begin{gathered} f\left(x\right)=0 \\ {x}^2+2x+1=0 \\ (x+1)(x+1)=0 \\ {(x+1)}^2=0 \end{gathered}$$

The point where the function intersect the x-axis are -

(x= -1)

Determine the point at which the function intersects the y-axis -

$$\begin{gathered} f\left(0\right)=0^2+2\times 0+1 \\ f\left(0\right)=1 \end{gathered}$$

The point at which the function intersect the y-axis is -  

$y=1$

Part (b). Step 1. Given data

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

Part (b). Step 2. To Find

Determine the domain and the range of the function.

Part (b). Step 3. Explanation

• The Domain is -   $\mathrm{ℝ}$
• And the range is -   $y\in [0, +\infty )$

Part (c). Step 1. Given data

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

Part (c). Step 2. To Find

Determine where the function is increasing and where it is decreasing.

Part (c). Step 3. Explanation

• Function increasing for -   $x\in [-1, +\infty )$
• Function decreasing for -   $x\in (-\infty , -1]$

Graph

The graph of function will be -