Chapter 10.9, Problem 32E

To determine

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

Answer to Problem 32E

(-3,9) and (2,4) are the vertices where the parabola and the straight line intersect

Explanation of Solution

Given:

  $y\ge x^2$

  $x+y\ge 6$

Calculation:

The given function

  $y\ge x^2$

  $x+y\ge 6$

Here, it is given that

  $x^2\le y$

  $x+y\ge 6$

Then, we take

  $x^2=y\dots \dots \mathrm{..}(1)$

The vertex of the parabola (0,0)

The equation of the straight line $x+y=6\dots \dots \mathrm{...}(2)$

  $\frac{x}{6}+\frac{y}{6}=1$

The line passes through points (6,0) and (0,6)

From equation (2), we have

  $y=6-x$

Substituting this expression into equation $(1),$ we get

  $x^2=6-x$

  $x^2+x-6=0$

  $(x+3)(x-2)=0$

  $x=-3$ and $x=2$

When $x=-3,$ then $y=6+3=9$

When $x=2,$ then $y=6-2=4$

(-3,9) and (2,4) are the vertices where parabola and the straight line intersect.

  

Conclusion:

Therefore, (-3,9) and (2,4) are the vertices where the parabola and the straight line intersect