Question

HELP!!! Sketch the graph of f based on the following characteristics:
- f(0) = f(2) = 0.
- f'(x) < 0 if x < 1.
- f'(x) = 0.
- f(x) > 0 if x > 1.
- f''(x) > 0.

Answer 1

Are there the graphs of in-equality

Answer 2

i bet it is a parabola with a positive leading coefficient
maybe make a quadratic

Answer 3

f(o) =f(2)=0 are x intercepts

Answer 4

this statement

- f'(x) = 0.
is a bit confusing
i think it is a typo

Answer 5

i bet it says \(f'(1)=0\)

Answer 6

Yeah, you are right! It's f'(1) = 0! My mistake! Sorry!

Answer 7

check \(f(x)=x^2-2x\) and see if it fits all your conditions

Answer 8

reason being that since the zeros are at \(x=0\) and \(x=2\) you know it looks like
\[f(x)=ax(x-2)=ax^2-2ax\]
the leading coefficient is positive, since the second derivative is \(2a\) which you are told is positive, and the first coordinate of the vertex is \(1\) which will give you \(a=1\) as well

Answer 9

YEAH! That's totally right! I was checking f(x) = x^2 - 2x, and it fits all the conditions! Thank you so much! I really appreciate it!

Answer 10

Also, this is what I got, too: f(x) = ax(x-2)=ax^2 - 2ax.

Answer 11

I thought it was a parabola, but I wasn't sure; now I am sure!