Q. 43 - 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 -  $f\left(x\right)=3x^2+6x+2$

Step-by-Step Solution

Answer

(a). 

• The graph is open upwards.
• The vertex of graph is - (-1, -1)
• 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  -  $y\in [-1, +\infty )$

(c).

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

Part (a). Step 1. Given data

Function -  $f\left(x\right)=3x^2+6x+2$

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

$$\begin{gathered} -\frac{b}{2a}=-\frac{6}{2\times 3} \\ -\frac{b}{2a}=-1 \end{gathered}$$

$$\begin{gathered} f(-\frac{b}{2a})=f(-1) \\ f(-\frac{b}{2a})=3\times {(-1)}^2+6\times (-1)+2 \\ f(-\frac{b}{2a})=3-6+2 \\ f(-\frac{b}{2a})=-1 \end{gathered}$$

Vertex is -   (-1, -1)

Determine the Axis of symmetric -

$$\begin{gathered} x=-\frac{b}{2a} \\ x=-\frac{6}{2\times 3} \\ x=-1 \end{gathered}$$

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 -

$$\begin{gathered} f\left(x\right)=0 \\ 3x^2+6x+2=0 \\ {x}_{1,2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a} \\ {x}_{1,2}=\frac{-6\pm \sqrt{6^2-4\times 3\times 2}}{2\times 3} \\ {x}_{1,2}=\frac{-6\pm 3.46}{6} \end{gathered}$$

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

$\mathbf{(}{\mathit{x}}_{\mathbf{1}}\mathbf{=}\mathbf{ }\mathbf{0}\mathbf{.}\mathbf{423}\mathbf{ }\mathit{a}\mathit{n}\mathit{d}\mathbf{ }{\mathit{x}}_{\mathbf{2}}\mathbf{=}\mathbf{ }\mathbf{1}\mathbf{.}\mathbf{577}\mathbf{)}\mathbf{ }$

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

$$\begin{gathered} f\left(0\right)=3\times 0^2+6\times 0+2 \\ f\left(0\right)=2 \end{gathered}$$

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

$y=2$

Part (b). Step 1. Given data

Function -  $f\left(x\right)=3x^2+6x+2$

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 -  $y\in [-1,+\infty )$

Part (c). Step 1. Given data

Function -  $f\left(x\right)=3x^2+6x+2$

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 -