Chapter 1.2, Problem 11E

Solution

Graph the function. Label the vertex and axis of symmetry.

Axis of symmetry (x) $=\frac{-b}{2a}=3$

Vertex $=(3,2)$

Given:

The given function $y=\frac{1}{2}{(x-3)}^2+2$

Concept Used:

• The vertex is the highest point if the parabola opens downward and the lowest point if the parabola opens upward.
• The axis of symmetry is the line that cuts the parabola into 2 matching halves and the vertex lies on the axis of symmetry.

Calculation:

The graph of the given function $y=\frac{1}{2}{(x-3)}^2+2$ is -

  

The given function $y=\frac{1}{2}{(x-3)}^2+2$

  $\begin{array}{c}$ y=\frac{1}{2}{(x-3)}^2+2$y=\frac{1}{2}(x^2+9-6x)+2\\ y=\frac{1}{2}{x}^2+\frac{9}{2}-3x+2\\ y=\frac{1}{2}{x}^2-3x+4.5+2\\ y=\frac{1}{2}{x}^2-3x+6.5\end{array}$

Here $a=\frac{1}{2},b=-3,c=6.5$

Axis of symmetry (x) $=\frac{-b}{2a}$

  $\begin{array}{l}=\frac{-(-3)}{2(\frac{1}{2})}\\ =\frac{3}{1}\\ =3\end{array}$

So Axis of symmetry (x) $=\frac{-b}{2a}=3$

Vertex $=\left(\frac{-b}{2a},y(\frac{-b}{2a})\right)$

  $\begin{array}{c}y\left(\frac{-b}{2a}\right)=y\left(3\right)\\ $ y=\frac{1}{2}{(x-3)}^2+2$$ y=\frac{1}{2}{(3-3)}^2+2$=0+2\\ =2\end{array}$

Vertex $=\left(\frac{-b}{2a},y(\frac{-b}{2a})\right)$

Vertex $=(3,2)$