Question

HELP ASAP PLEASE

Answer 1

The correct method:

1. Find out dy/dx
2. Put dy/dx = 0 and solve for x
3. Find out d^2 y/dx^2 ...
4. Put the values of x you obtained from step 2 in expressions from step 3
5. If d^2 y/dx^2 < 0 => Maximum at that point; if d^2 y/dx^2 > 0 => Minimum at that point... I won't go deep into d^2 y/dx^2 = 0

Answer 2

Would the answer be C?

Answer 3

dy/dx = -2x + 4 = 0 => x = 2

d^2 y/dx^2 = -2

At x = 2, d^2 y/dx^2 < 0 => Maximum

y @ x = 2 is -4 + 8 - 4 = 0

So you are partially correct.. the correct answer is A

Answer 4

Okay, thank you! Can you help me with a similar question, please?

Answer 5

Sure

Answer 6

What do you think?

Answer 7

What is dy/dx here?

Answer 8

I think B

Answer 9

Nope

Answer 10

How did you get it?

Answer 11

Ohh wait, I was looking at the question wrong!

Answer 12

Can you give me a hint?

Answer 13

Yeah sure

First find out dy/dx... You will get dy/dx = x + 8

Put it equal to 0, find out the value of x

Then find out d^2y/dx^2

Answer 14

Ohh would it be C? Because 8-8=0?

Answer 15

Wait what is d^2y/dx^2 ?

Answer 16

I'm not sure.

Answer 17

Okay can you find out x from dy/dx = 0 ?

Answer 18

No, I'm confused on what dy/dx is

Answer 19

You have done calculus right?

Answer 20

No. I'm only in Algebra 2.

Answer 21

Oops!! My mistake then...

Answer 22

Okay so we have another option...

Answer 23

Now the above was the graph when a is positive


When a is -ve the graph becomes: