Q. 44 - 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 -  $f\left(x\right)=2x^2+5x+3$

Step-by-Step Solution

Answer

(a). 

• The graph is open upwards.
• The vertex of graph is - (-1.25, -0.125)
• 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  -  $y\in [-0.125, +\infty )$

(c).

• Function increasing for -  $x\in [-1.25, +\infty )$
• Function decreasing for -  $x\in (-\infty , -1.25]$

Part (a). Step 1. Given data

Function -  $f\left(x\right)=2x^2+5x+3$

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.

$$\begin{gathered} -\frac{b}{2a}=-\frac{5}{2\times 2} \\ -\frac{b}{2a}=-1.25 \end{gathered}$$

$$\begin{gathered} f(-\frac{b}{2a})=f(-1.25) \\ f(-\frac{b}{2a})=2\times {(-1.25)}^2+5\times (-1.25)+3 \\ f(-\frac{b}{2a})=3.125-6.25+3 \\ f(-\frac{b}{2a})=-0.125 \end{gathered}$$

 

Vertex is -   (-1.25, -0.125)

Determine the Axis of symmetric -

$$\begin{gathered} x=-\frac{b}{2a} \\ x=-\frac{5}{2\times 2} \\ x=-1.25 \end{gathered}$$

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 -

$$\begin{gathered} f\left(x\right)=0 \\ 2x^2+5x+3=0 \\ (x+1)(x+\frac{3}{2})=0 \end{gathered}$$

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

(x= -1 and x= -1.5)

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

$$\begin{gathered} f\left(0\right)=2\times 0^2+5\times 0+3 \\ f\left(0\right)=3 \end{gathered}$$

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

$y=3$

Part (b). Step 1. Given data

Function -  $f\left(x\right)=2x^2+5x+3$

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 -  $y\in [-0.125, +\infty )$

Part (c). Step 1. Given data

Function -  $f\left(x\right)=2x^2+5x+3$

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  -   $x\in [-1.25, +\infty )$
• Function decreasing for  -   $x\in (-\infty , -1.25]$

Graph

The graph of function will be -