Answer:
1
Step-by-step explanation:
Sometimes the easiest way to answer a question like this one is to graph it.
The point has to be closest to 1,5. It can also be 1,5
So the answer is a=1
Answer:
1
Step-by-step explanation:
Let's plug the point into both inequalities.
[tex]5 \ge (a-2)^2+4[/tex]
[tex]5 \le -(a-2)^2+6[/tex]
We will solve both and see what values of [tex]a[/tex] they both have in common and then determine the least of that set.
Let's start with the first:
[tex]5 \ge (a-2)^2+4[/tex]
Subtract 4 on both sides:
[tex]1 \ge (a-2)^2[/tex]
We are looking for numbers whose squares are less than 1. Those numbers are between -1 and 1.
So this gives me:
[tex]-1\le a-2 \le 1[/tex]
Add 2 on all sides:
[tex]1 \le a \le 3[/tex]
Now let's solve the other inequality:
[tex]5 \le -(a-2)^2+6[/tex]
Subtract 6 on both sides:
[tex]-1 \le -(a-2)^2[/tex]
Multiply both sides by -1:
[tex] 1 \ge (a-2)^2[/tex]
We are already solved this inequality above.
So the solution for [tex]a[/tex] is [tex]1 \le a \le 3[/tex].
The smallest number of that set is 1.
