Question

can some one graph this please??
ill award and fan !
it is an upside-down parabola with a vertex at x=1 (third charateristic) and zeros at x=0 and x=2 (first charateristic) and the second says that the slope of the tangent line is positive when x<1 so the graph rises from left to right until it gets to x=1 and the fourth says that the slope is negative meaning it falls from left to right after x=1 making the graph an upside-down parabola

Answer 1

The vertex is (1,0)? and it has two other zeros? Or is it just a plain old parabola with the vertex at (1,1)?

Answer 2

@ovenmitt12 the original question is this Sketch the graph of a function f(x) having the following characteristics...?
f (0) = f (2) = 0
f ' (x) > 0 if x < 1
f ' (1) = 0
f ' (x) < 0 if x > 1
f " (x) < 0

Answer 3

i thought that would be what the graph would look like

Answer 4

I think you have it.... It would be a parabola with:

vertex (1,1) P(0,0) and P(2,0)

and with f''(x) < 0 it would be concave down....

Answer 5

My turn... What are the zeros of x^3-4x+1?

Answer 6

i understand it i just don't write get how im suppose to graph that :(

Answer 7

I think you just have to graph the original function that would have been a parabola... I don't think you would have to graph any of the derivative or the second deriv.

Answer 8

can you draw that out please,because what the graph is suppose to look like is were im lost