Question

Find the maximum value of |x^2+6x-7| for |x+3|≤4.

Answer 1

I keep getting 7 as an answer but the correct answer is 16. Help

Answer 2

well first we can set some kind of inequality for x

\[-7 \le x \le 1\]

Answer 3

i got the answer but im not sure if theres a simpler way to explain it...

Answer 4

just guessed and checked.

Answer 5

Do you know how to differentiate?

Answer 6

Or actually we could just use algebra if you don't know how to do that?

Answer 7

So tyler solve the |x+3|<=4
and got -7<=x<=1

Well |x^2+6x-7|=-(x^2+6x-7) when x is in the interval [-7,1]

Find the x-coordinate of the vertex
and then plug that into |x^2+6x-7| to get your answer.

Answer 8

And I'm not about finding the vertex of the parabola that is f(x)=-(x^2+6x-7)

Answer 9

You can do that using calculus or algebra.

Answer 10

Please say something so I can know you are there or not.
It will also be helpful to me if you said you understand or don't understand and said what you didn't understand.

Answer 11

I'm taking AP Cal the following semester.

Answer 12

i need to solve via algebra

Answer 13

Ok do you know how to find the vertex of a parabola?

Answer 14

-b/2a?

Answer 15

-b/(2a) will give you the x-coordinate

Answer 16

so what would that be?

Answer 17

-3

Answer 18

Ok now plug that into |x^2+6x-7|

Answer 19

OOHHHH

Answer 20

Thank you!

Answer 21

We found the vertex of that parabola because that would have given us the max value in the interval [-7,1] of that absolute value function.

Answer 22

and no problem

Answer 23

so wait, would it be two parabolas bc its absolute value? positive parabola has a minimum right?

Answer 24

@myininaya

Answer 25

The absolute value function looks like this


Basically you graph y=x^2+6x-7 first

And since we have y=|x^2+6x-7| this means you take every number from y=x^2+6x-7 and make sure it is positive if it is not take the opposite of it
That means we would take the part that is underneath the x-axis and flip it about the x-axis
giving us the first graph I mentioned

Also if you wanted to find the piecewise function that absolute value function is equal to in symbols instead of drawings

Then you would need to know this definition



So we have
|x^2+6x-7|=x^2+6x-7 if x^2+6x-7>=0
x^2+6x-7>=0 when x<-7 or x>1
|x^2+6x-7|=-(x^2+6x-7) if x^2+6x-7<0
x^2+6x-7<0 when -7<x<1

Or you can even rewrite your absolute value definition
you could have said
|f(x)|=-f(x) if f(x)<=0


So you have the |x^2+6x-7|=-(x^2+6x-7) if -7<=x<=1









Anyways to answer your question you are only concerned with happens between -7 and 1.
You only have the one parabola.