Chapter 1.2, Problem 10E

Solution

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

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

Vertex $=(-2,1)$

Given:

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

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 $y=\frac{-1}{4}{(x+2)}^2+1$ is -

  

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

  $\begin{array}{c}$ y=\frac{-1}{4}{(x+2)}^2+1$y=\frac{-1}{4}(x^2+4+4x)+1\\ y=\frac{-1}{4}{x}^2-1-x+1\\ y=\frac{-1}{4}{x}^2-x\end{array}$

Here $a=\frac{-1}{4},b=-1,c=0$

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

  $\begin{array}{l}=\frac{-(-1)}{2(\frac{-1}{4})}\\ =\frac{1(4)}{-2}\\ =\frac{4}{-2}\\ =-2\end{array}$

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

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

  $\begin{array}{c}y\left(\frac{-b}{2a}\right)=y\left(-2\right)\\ $ y=\frac{-1}{4}{(x+2)}^2+1$$ y=\frac{-1}{4}{(-2+2)}^2+1$=0+1\\ =1\end{array}$

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

Vertex $=(-2,1)$