Chapter 1.6, Problem 48E

Solution

The graph G that is obtained by sequence of transformation in graph of y.

The graph G is $y=3{\left(x-4\right)}^2$ .

Given:

The graph of $y=x^2$ is shifted by $4$ units right, then a vertical stretch by a factor $3$ .

Concept Used:

The graph of $y=f\left(x\right)$ is translated or shifted horizontally left or right according as:

1) If $y=f\left(x-c\right)$ , then graph shifted by c units right from $y=f\left(x\right)$ .

2) If $y=f\left(x+c\right)$ , then graph shifted by c units left from $y=f\left(x\right)$ .

And, the graph of $y=f\left(x\right)$ is translated or shifted vertically up or down according as:

1) If $y=f\left(x\right)+c$ , then graph shifted by c units up from $y=f\left(x\right)$ .

2) If $y=f\left(x\right)-c$ , then graph shifted by c units down from $y=f\left(x\right)$ .

The transformation of horizontal stretches or shrinks from $y=f\left(x\right)$ is as follows:

The graph of $y=f\left(\frac{x}{c}\right)$ transform to horizontal stretches by factor c , when $c>1$ and horizontal shrinks by factor c when $c<1$ .

The transformation of vertical stretches or shrinks from $y=f\left(x\right)$ is as follows:

The graph of $y=c\cdot f\left(x\right)$ transform to vertical stretches by factor c , when $c>1$ and shrinks by factor c when $c<1$ .

Calculation:

Consider the sequence of transformation,

The graph of function $y=x^2$ is shifted by $4$ units right,

So x is replace by $\left(x-4\right)$ , that is, $y={\left(x-4\right)}^2$ .

Thus, the equation $y=x^2$ transform to $y={\left(x-4\right)}^2$ .

And, in the next transformation, the graph of $y={\left(x-4\right)}^2$ is stretches vertically by factor $3$ ,

So $3$ is multiplied to the function, that is $y=3{\left(x-4\right)}^2$ as $3>1$ .

Hence, the function $y={\left(x-4\right)}^2$ transform to $y=3{\left(x-4\right)}^2$ .

Conclusion The equation $y=x^2$ transform to $y=3{\left(x-4\right)}^2$ when graph of $y=x^2$ is shifted by $4$ units right, then a vertical stretch by a factor $3$ .