Chapter 1.2, Problem 3GP

Solution

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

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

Vertex $=(3,-4)$

Given:

The given function $f(x)=\frac{1}{2}{(x-3)}^2-4$

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 $f(x)=\frac{1}{2}{(x-3)}^2-4$ is -

  

The given function $f(x)=\frac{1}{2}{(x-3)}^2-4$

  $\begin{array}{c}$ f(x)=\frac{1}{2}{(x-3)}^2-4$f(x)=\frac{1}{2}(x^2+9-6x)-4\\ f(x)=\frac{x^2}{2}+\frac{9}{2}-\frac{6x}{2}-4\\ f(x)=\frac{x^2}{2}+\frac{9}{2}-3x-4\\ f(x)=\frac{x^2}{2}-3x-4+4.5\\ f(x)=0.5x^2-3x+0.5\end{array}$

Here $a=0.5,b=-3,c=0.5$

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

  $\begin{array}{l}=\frac{-(-3)}{2(0.5)}\\ =\frac{3}{1}\\ =3\end{array}$

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

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

  $\begin{array}{c}f\left(\frac{-b}{2a}\right)=f\left(3\right)\\ $ f(x)=\frac{1}{2}{(x-3)}^2-4$$ f(3)=\frac{1}{2}{(3-3)}^2-4$f(3)=-4\end{array}$

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

Vertex $=(3,-4)$