Chapter 1.7, Problem 59E

Solution

a.

To graph: The functions $y=x^2+2x,y=x^2+4x,y=x^2-6x,y={\left(x+1\right)}^2,y={\left(x+2\right)}^2,y={\left(x-3\right)}^2$

Graphs are as follows.

  

  

  

  

Given:

Functions are $y=x^2+2x,y=x^2+4x,y=x^2-6x,y={\left(x+1\right)}^2,y={\left(x+2\right)}^2,y={\left(x-3\right)}^2$

Graph:

Consider the function $y=x^2+2x$ .

Substitute some values for $x$ , say $-4$ , and evaluate $y$ .

  $\begin{array}{c}y={\left(-4\right)}^2+2\left(-4\right)\\ =16-8\\ =8\end{array}$

Organize the results in a table.

  $\begin{array}{cccccc}x& -4& -2& -1& 1& 2\\ y& 8& 0& -1& 3& 8\end{array}$

Plot the points on a coordinate plane and connect them with a smooth curve.

  

Similarly, graph $y={\left(x+1\right)}^2$ on the same coordinate plane.

  

Graph the second pair of functions on a coordinate plane.

  

Graph the third pair of functions on a coordinate plane.

  

b.

To compare: The graphs of $y=x^2+bx$ and $y={\left(x+\frac{b}{2}\right)}^2$ .

The graphs of $y={\left(x+\frac{b}{2}\right)}^2$ are narrower than the graphs of $y=x^2+bx$

Given:

Equations are $y=x^2+bx,y={\left(x+\frac{b}{2}\right)}^2$

Calculation:

On comparing each pair of graphs, it is observed that the graphs of $y={\left(x+\frac{b}{2}\right)}^2$ are narrower than the graphs of $y=x^2+bx$ with their vertices shifted to the $x-$ axis. On completing the square of $y=x^2+bx$ , the new equation is $y={\left(x+\frac{b}{2}\right)}^2$ . Therefore, both the graphs will be the same.