Chapter 1.4, Problem 77E

a.

To determine

To graph a function using plotting points or translating parent function.

Explanation of Solution

Given:

Function: $f(x)=3x^2-4x+1$

Formula used:

Quadratic equation: $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Calculation:

Substituting the values of given equation in the above formula,

  $\begin{array}{l}x=\frac{-(-4)\pm \sqrt{{(-4)}^2-4(3)(1)}}{2(3)}\\ x=\frac{4\pm \sqrt{16-12}}{6}\\ x=\frac{4\pm \sqrt{4}}{6}\\ x=\frac{4\pm 2}{6}\\ x=\frac{6}{6},\frac{2}{6}\\ x=1,0.33\end{array}$

Substitute the value of $x$ in $f(x)=3x^2-4x+1$ to find $y=f(x)$

When $x=1,$

  $\begin{array}{c}$ f(x)=3{(1)}^2-4(1)+1$=3-4+1\\ =0\end{array}$

The point is $(1,0)$

When $x=0.33,$

  $\begin{array}{c}$ f(x)=3{(0.33)}^2-4(0.33)+1$=0.3267-1.32+1\\ =0.0067\end{array}$

The other point is (1,0.0067)

Graph:

Plot the points in the graph as shown below:

  

Interpretation:

Looking at the function given, the structure of the function is similar to that of the general formula for finding quadratic equations. So, in such cases plotting the points is better to use.

b.

To determine

To graph a function using plotting points or translating parent function.

Explanation of Solution

Given:

Function: $f(x)=2{(x-1)}^2-6$

Calculation:

Using translating parent function method,

The general equation for horizontal stretch is $y={(b(x-h))}^2+k$

Here, the given function is similar to the above equation.

So, the new vertex of the function is $(1,-6)$ [one unit to the right and 6 units downward]

Therefore, the horizontal stretch is $\frac{1}{2}$ (reciprocal of 2, since the function is horizontal)

So the vertex of the function is $(1,-6)$

Graph:

  

Interpretation:

Looking at the function given, the structure of the function is similar to that of the horizontal stretch. So, in such cases translating parent function is better to use.