Chapter 2, Problem 63RE

To determine

To sketch: the graph of the function. Also state the domain, intercepts and range of the graph.

Answer to Problem 63RE

The intercepts are $\left(0,3\right)$ .

The interval $\left[-\infty ,\infty \right]$ .

The intercepts are $\left(0,3\right)$ , domain is the set of all real numbers and range is $\left\{y|y\ge 2\right\}$ or the interval $\left[2,\infty \right)$ .

Explanation of Solution

Given information:

Given function

  $h\left(x\right)={\left(x-1\right)}^2+2$

Calculation:

Graph the basic function $h\left(x\right)=x^2$ .

  

Shift the graph of $h\left(x\right)=x^2$ , horizontally right 1 unit to obtain $h\left(x\right)={\left(x-1\right)}^2$

  

Shift the graph of $h\left(x\right)={\left(x-1\right)}^2$ vertically up by 2 unit to get the graph of $h\left(x\right)={\left(x-1\right)}^2+2$ .

  

Intercepts are the points where the graph touches or crosses the coordinate axes. In $h\left(x\right)={\left(x-1\right)}^2+2$ , the graph touches the coordinate axes at $\left(0,3\right)$

So, the intercepts are $\left(0,3\right)$ .

The points on the graph have $x$ coordinates between $-\infty$ and $\infty$ , and for each number $x$ between $-\infty$ and $\infty$ , there is a point $\left(x,f\left(x\right)\right)$ on the graph. So, the interval $\left[-\infty ,\infty \right]$ .

The points on the graph have $y$ coordinates between 2 and $\infty$ , and for each number $y$ , there is at least one number $x$ in the domain.

So, the range of $F\left(x\right)$ is $\left\{y|y\ge 2\right\}$ or the interval $\left[2,\infty \right)$ .

Therefore, the intercepts are $\left(0,3\right)$ , domain is the set of all real numbers and range is $\left\{y|y\ge 2\right\}$ or the interval $\left[2,\infty \right)$ .