Chapter 1.1, Problem 22E

Solution

To graph: A function and label the vertex and axis of symmetry.

Given information:

The function is $y\left(x\right)=3x^2-6x+4$ .

Graph:

The steps followed for graphing the function of the form $y=a{x}^2+bx+c$ are:

Step-1: Find coefficients $a,b,c$ . Here, $a=3,b=-6,c=4$ .

Step-2: Find vertex $\left(x,y\right)$ where $,$ that is

  $\begin{array}{c}x=-\frac{-6}{2\cdot 3}=+\frac{6}{6}\\ =1,\\ y=y\left(-\frac{b}{2a}\right)=y\left(1\right)\leftarrow \text{substitute}x=1\text{in}y\left(x\right)\\ =3{\left(1\right)}^2-6\left(1\right)+4\\ =3-6+4\\ =1\end{array}$

Thus, the vertex of the parabola represented by the function $y\left(x\right)=3x^2-4x+4$ is $\left(1,1\right)$ .

Step-3: The equation of the axis of symmetry (line) is $x=-\frac{b}{2a}$ . Here, the axis of symmetry is $x=1$ .

Step-4: Find other points $\left(a,y\left(a\right)\right)$ on the curve.

Step-5: Plot axis of symmetry, vertex and other points. The points table and the graph of the function is shown in Fig. (1).

Since $a=3>0$ , therefore the parabola opens upward.

  

Interpretation:

The axis of symmetry divides the area of the region bounded by the curve into two equal parts.

The vertex of the graph of a function is a point where curve bends making this point a lowest or highest point in the curve where maxima or minima of the function exists.