Chapter 8, Problem 23CR

To determine

Graph the quadratic function. Label the vertex, axis of symmetry, and $\mathrm{x}$ -intercepts. Describe the domain and range of the function.

Answer to Problem 23CR

Vertex : $(-2,3)$

Axis of Symmetry : x = -2

x -intercepts are (-3,0) and (-1,0)

  $\mathrm{D}=\{\mathrm{x}:\mathrm{x}\in \mathrm{ℝ}\}$

  $\mathrm{R}=\{\mathrm{y}:\mathrm{y}\in \mathrm{ℝ}\text{ and }\mathrm{y}\le 3\}$

Explanation of Solution

Given Information:

Given function is :

  $\mathrm{f}(\mathrm{x})=-3(\mathrm{x}+3)(\mathrm{x}+1)$

Concept Used :

• Quadratic Functions:
• Quadratic functions are non-linear and are of the form:

  $\mathrm{f}(\mathrm{x})=\mathrm{a}{\mathrm{x}}^2+\mathrm{b}\mathrm{x}+\mathrm{c};\mathrm{a}\ne 0$

This is the standard form of Quadratic function.

• Shape of the graph of a Quadratic function is called Parabola.
• Parabolas are symmetric about a central line called Axis Of Symmetry.
• The axis of symmetry intersects a parabola only at one point , called the Vertex.
• Axis of Symmetry: Passes through the vertex and divides the parabola into two congruent halves.

  $\mathrm{x}=-\frac{\mathrm{b}}{2\mathrm{a}}$

• If a < 0 ,
• The graph of $\mathrm{a}{\mathrm{x}}^2+\mathrm{b}\mathrm{x}+\mathrm{c}$ opens downward.
• The highest point on the graph is the maximum.
• The maximum or the minimum is the vertex.

• y -intercept = The y coordinate of the point at which the graph intersects y -axis , x = 0.
• x -intercept = The x coordinate of the point at which the graph intersects x -axis , y = 0 .

Calculation:

• Convert the given function into equation : $\mathrm{y}=-3(\mathrm{x}+3)(\mathrm{x}+1)$
• In the equation

$\begin{array}{l}\mathrm{y}=-3(\mathrm{x}+3)(\mathrm{x}+1)\\ \Rightarrow \mathrm{y}=-3({\mathrm{x}}^2+\mathrm{x}+3\mathrm{x}+3)\\ \Rightarrow \mathrm{y}=-3({\mathrm{x}}^2+4\mathrm{x}+3)\\ \Rightarrow \mathrm{y}=-3{\mathrm{x}}^2-12\mathrm{x}-9\end{array}$

comparing it with standard form of quadratic equation , we have a = -3 , b = -12 , c = -9 .

• Since a = -3 < 0 , so , the parabola opens downward and the maximum point is vertex.
• Vertex :
• The x -coordinate of the vertex is $\mathrm{x}=-\frac{\mathrm{b}}{2\mathrm{a}}=-\frac{-(-12)}{2(-3)}=-2$
• Substitute x = -2 in the given function to find the y -coordinate of the vertex :

$\begin{array}{l}\mathrm{y}=-3{\mathrm{x}}^2-12\mathrm{x}-9\\ \mathrm{y}=-3{(}^{-}-12(-2)-\mathrm{9.........................}[\mathrm{x}=-2]\\ \mathrm{y}=-12+24-9\\ \mathrm{y}=3\end{array}$

• So, the vertex is at (-2,3)
• Find Axis of Symmetry :

The Axis of Symmetry is

  $\begin{array}{l}\mathrm{x}=-\frac{\mathrm{b}}{2\mathrm{a}}=-2\\ \Rightarrow \mathrm{x}=-2\end{array}$

• Find y -intercept :

The y -coordinate of the point at which the graph cuts the y -axis is the y -intercept.

Hence , substituting x =0 in the equation $\mathrm{y}=-3{\mathrm{x}}^2-12\mathrm{x}-9$ , we will get the y -intercept.

  $\begin{array}{l}\mathrm{y}=-3{\mathrm{x}}^2-12\mathrm{x}-9\\ \Rightarrow \mathrm{y}=-3{(}^0-12(0)-\mathrm{9.............}[\mathrm{x}=0]\\ \Rightarrow \mathrm{y}=-9\end{array}$

So, y -intercept = (0,-9)

• Find x -intercept :

The x -coordinate of the point at which the graph cuts the x -axis is the x -intercept.

Hence , substituting y =0 in the equation $\mathrm{y}=-3{\mathrm{x}}^2-12\mathrm{x}-9$ , we will get the x -intercept.

  $\begin{array}{l}\mathrm{y}=-3{\mathrm{x}}^2-12\mathrm{x}-9\\ \Rightarrow 0=-3{\mathrm{x}}^2-12\mathrm{x}-\mathrm{9.............}[\mathrm{y}=0]\\ \Rightarrow 0={\mathrm{x}}^2+4\mathrm{x}+\mathrm{3...................}[\text{divide each side by -3]}\\ \Rightarrow {\mathrm{x}}^2+3\mathrm{x}+\mathrm{x}+3=\mathrm{0.......................}[\mathrm{f}\mathrm{a}\mathrm{c}\mathrm{t}\mathrm{o}\mathrm{r}\mathrm{i}\mathrm{z}\mathrm{e}\text{ middle term]}\\ \Rightarrow (\mathrm{x}+1)(\mathrm{x}+3)=0\\ \Rightarrow \mathrm{x}=-1\text{ and }\mathrm{x}=-3\end{array}$

So, x -intercepts are (-3,0) and (-1,0).

Use the above information to graph the equation $\mathrm{y}=-3(\mathrm{x}+3)(\mathrm{x}+1)$:

  

Clearly , since the function takes values from whole real line, so , the domain is :

  $\mathrm{D}=\{\mathrm{x}:\mathrm{x}\in \mathrm{ℝ}\}$

Since the graph has maximum at vertex (-2,3), so the range is :

  $\mathrm{R}=\{\mathrm{y}:\mathrm{y}\in \mathrm{ℝ}\text{ and }\mathrm{y}\le 3\}$