Chapter 3, Problem 1RE

To determine

To graph:

Quadratic function $y={(x-1)}^2+2$.

Explanation of Solution

Given:

$y={(x-1)}^2+2$

Graph :

1. Rearrange the equation as follows

$\begin{array}{l}y={(}^x+2\\ =x^2-2x+3\end{array}$

Coefficient of the $x^2$ term is positive.

Therefore, the parabola opens up

2. At vertex $\frac{dy}{dx}=0$

$\begin{array}{l}y={(}^x+2\\ \frac{dy}{dx}=2(x-1)\end{array}$

When $x=1\to \frac{dy}{dx}=0$

Therefore, $x=1,y=2$ is the vertex.

3. For a quadratic equation, the axis is parallel to y axis and it crosses the vertex.

Therefore, axis is $x=1$

4. When parabola intersects the $x$ axis the $y$ coordinate will be $0$

Therefore,

$\begin{array}{l}0={(}^x+2\\ ={(}^x+2\\ {(}^x\text{ =-2}\end{array}$

Solutions for $x$ become complex. Therefore, parabola does not intercept $x$ axis

5. When parabola intersects the $y$ axis the $x$ coordinate will be $0$

Therefore,

$\begin{array}{l}y={(}^x+2\\ ={(}^0+2\\ \text{ =3}\end{array}$

$y$ Intercept is $x=0,y=3$

6. $\frac{dy}{dx}=2(x-1)$

When $x<1\to \frac{dy}{dx}<0$

When $x>1\to \frac{dy}{dx}>0$

When $x=1\to \frac{dy}{dx}=0$

Therefore, the increasing region is $x\subset (1,\infty )$ and the decreasing region is $x\subset (-\infty ,1)$

The graph can be drawn as follows.

Interpretation :

We can start drawing by pointing the vertex and $x$ intercept and $y$. Then complete the drawing by considering it is opens up and increasing and decreasing regions. Additionally use the fact that the parabola is symmetric around its axis.