Chapter 8, Problem 12CR

To determine

Tograph:The function $y=\frac{1}{2}{x}^2-6x+10$ and describe its domain and range.

Answer to Problem 12CR

The domain and range of $y=\frac{1}{2}{\left(x-6\right)}^2-8$ are intervals $\left(-\infty ,\infty \right)$ and $\left[-8,\infty \right)$ respectively.

Explanation of Solution

Given information:

The given function to graph is $y=\frac{1}{2}{x}^2-6x+10$ , of which domain and range is to be determined.

Formulaused:

The standard form of the quadratic function is $f\left(x\right)=a{\left(x-h\right)}^2+k$ whose vertex is $\left(h,k\right)$ , and the domain is set of all real numbers or $\left(-\infty ,\infty \right)$ and range $y\ge k$.

Graphs:

The standard form of given function $y=\frac{1}{2}{x}^2-6x+10$ is

  $y=\frac{1}{2}\left(x^2-12x\right)+10$

  $y=\frac{1}{2}\left(x^2-12x+36-36\right)+10$

  $y=\frac{1}{2}\left(x^2-12x+36\right)-18+10$

  $y=\frac{1}{2}{\left(x-6\right)}^2-8$

This is a parabola with vertex $\left(6,-8\right)$ ,

The graphs of function $y=\frac{1}{2}{\left(x-6\right)}^2-8$ is shown in Figure-1 here.

  

From the function, $y=\frac{1}{2}{\left(x-6\right)}^2-8$ and also from its graph, it is clear that for each real number ‘ x’ there exists image $y=\frac{1}{2}{\left(x-6\right)}^2-8$ .

Hence, the domain of function $y=\frac{1}{2}{\left(x-6\right)}^2-8$ is the set of all real numbers or $\left(-\infty ,\infty \right)$.

Since, the minimum (at the vertex) of $y=\frac{1}{2}{\left(x-6\right)}^2-8$ is $-8$ , i.e. $y\ge -8$ .

Thus, the range of the function $y=\frac{1}{2}{\left(x-6\right)}^2-8$ is $y\ge -8$ or

Range of $y=\left(-8,\infty \right]$.

Conclusion:

The domain and range of $y=\frac{1}{2}{\left(x-6\right)}^2-8$ are intervals $\left(-\infty ,\infty \right)$ and $\left[-8,\infty \right)$ respectively.