Chapter 10.9, Problem 31E

To determine

To find: the coordinates of all the vertices and determine whether the solution set bounds or not.

Answer to Problem 31E

Therefore $\left(\text{-3,0}\right)\text{ and }\left(\text{2,5}\right)$ are the vertices

Explanation of Solution

Given:

  $y<9-x^2$

  $y\ge x+3$

Calculation:

The given function

  $y<9-x^2$

  $y\ge x+3$

Here, we take

  $y=9-x^2\dots \dots \dots .(1)$

  $x^2=9-y$

  $x^2=-(y-9)$

The vertex of the parabola $\left(0,9\right)$

The equation of the straight line $y=x+3\dots \dots \dots .(2)$

  $x-y=-3$

  $\frac{x}{-3}+\frac{y}{3}=1$

Then the line passes through points $\left(\text{-3,0}\right)\text{ and }\left(\text{0,3}\right)$

From equation (2), we have

  $y=x+3$

Substituting this expression into equation (1), we get

  $x+3=9-x^2$

  $x=6-x^2$

  $x^2+x-6=0$

  $(x+3)(x-2)=0$

  $x=-3$ and $x=2$

When $x=-3,$ then $y=0$

When $x=2,$ then $y=5$

Therefore, $\left(\text{-3,0}\right)\text{ and }\left(\text{2,5}\right)$ are the vertices where the parabola and the straight line intersect

  

Conclusion:

Therefore $\left(\text{-3,0}\right)\text{ and }\left(\text{2,5}\right)$ are the vertices