Chapter 7.5, Problem 8WE

To determine

To graph each given equation. Label the vertex and axis of symmetry. Find all intercepts.

Explanation of Solution

Given information :

Given equation $y-5=-{\left(x+2\right)}^2$

We know that the graph of the equation $y-k=a{\left(x-h\right)}^2$ opens upwards when $a>0$ and downwards when $a<0$ . Higher the values of $a$ , narrower the graph will be. Further if $h>0$ then the graph will slide to its right whereas it will shift on left side if the value of $h<0$ .

When $k>0$ the graph slides upwards and when $k<0$ then graph slides downwards.

Here the value of $a=-1$ hence the graph of parabola opens downwards.

Further $k=5$ hence the parabola will shift 5 units upwards.

And $h=-2$ hence the graph will shift 2 units to the left.

Answer for given equation is,

  

Axis of symmetry is $x=-2$

To find $x$ intercepts we put $x=0$

We get,

  $\begin{array}{l}y-5=-4\\ y=1\end{array}$

Similarly put $y=0$

We get,

  $\begin{array}{l}-5=-{\left(x+2\right)}^2\\ 5={\left(x+2\right)}^2\\ x+2=\pm \sqrt{5}\\ x=-2\pm \sqrt{5}\\ x=0.236,-4.236\end{array}$