Question

SAT Math Question (Thanks!)

y is greater than or equal to (x-2)^2 + 4
y is less than or equal to -(x-2)^2 + 6
The ordered pair (a,5), where a is a constant, satisfies the system of inequalities above. What is the least possible value of a?

Answer 1

1

Explanation:

Let's plug the point into both inequalities.

`5 \ge (a-2)^2+4`

`5 \le -(a-2)^2+6`

We will solve both and see what values of
`a` they both have in common and then determine the least of that set.

Let's start with the first:

`5 \ge (a-2)^2+4`

Subtract 4 on both sides:

`1 \ge (a-2)^2`

We are looking for numbers whose squares are less than 1. Those numbers are between -1 and 1.

So this gives me:

`-1\le a-2 \le 1`

Add 2 on all sides:

`1 \le a \le 3`

Now let's solve the other inequality:

`5 \le -(a-2)^2+6`

Subtract 6 on both sides:

`-1 \le -(a-2)^2`

Multiply both sides by -1:

`1 \ge (a-2)^2`

We are already solved this inequality above.

So the solution for
`a` is
`1 \le a \le 3`.

The smallest number of that set is 1.

Answer 2

1

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