Chapter 1.1, Problem 26E

Solution

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

Given information:

The function is $f\left(x\right)=-6x^2-4x-5$ .

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, $a=-6,b=-4,c=-5$ .

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

  $\begin{array}{c}x=-\frac{-4}{2\cdot \left(-6\right)}\\ =-1/3\\ y=f\left(-\frac{b}{2a}\right)=f\left(-1/3\right)\leftarrow \text{substitute}x=-1/3\text{in}y\left(x\right)\\ =-6{\left(-1/3\right)}^2-4\left(-1/3\right)-5\\ =-\frac{2}{3}+\frac{4}{3}-5\\ =-13/3\end{array}$

Thus, the vertex of parabola represented by the function v is $\left(-1/3,-13/3\right)$ .

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

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 $a=-1<0$ , therefore the parabola opens down.

  

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.