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# visualising graphs watch

1. hi guys

the correct answer is E. i've used desmos to have a little play around with what this graph could look like as the values of a, b and c change, and i can definitely see how E has to be correct. however, i'm still confused as to how i would be able to deduce this using only a pen, paper, and my own mind.

i would appreciate any insight please thanks
2. (Original post by ashaxo99)
hi guys

the correct answer is E. i've used desmos to have a little play around with what this graph could look like as the values of a, b and c change, and i can definitely see how E has to be correct. however, i'm still confused as to how i would be able to deduce this using only a pen, paper, and my own mind.

i would appreciate any insight please thanks

From which we can see as x goes to infinity, so does y, so "a", the dominant term must be positive.

Then work out dy/dx. One root will be x=0, and the other must be greater than zero, since the turning point is in the 4th quadrant. Which gives b < 0. (details left for you to work out.)

If you work out y for the second stationary point, then "c" being negative is sufficient for y to be negative at that point. (again details left for you to work out).

Also, note the wording of the question; a sufficient condition. It's not neccsarily required that c is negative. If it's positive, then the turning point may or may not be in the 4th quadrant. But if it's negative, then that's sufficient to force the turning point to be in the 4th quadrant.

Edit: HeadmasterCid's method below of using the second derivative (for max at x=0) is a quicker way for determining the sign of b.
3. I would think (having not had a good go at this yet) that you will need to differentiate twice. Finding the first derivative will allow you to find an x value of the minimum in terms of a and b (c will differentiate to zero), and you can use this to find a y value in terms of a, b and c. You know that this x value needs to be greater than zero, and the y value be less than zero. There are two inequalities already.

Next you can use the fact that that point is a minimum with the second derivative to find another inequality, and also the maximum at x=0 to find a fourth inequality (although they should be the same).

That should hopefully eliminate 7/8 options, and will give you an answer!

UPDATE ghostwalker's help is much better than mine - I should have actually put pen to paper!
4. Interesting question.

The position of the cancavities of the graph should indicate "a".

By using the first derivative, this should indicate "b", (as the others above me have stated).

As ghostwalker stated, for "c", see what happens when "c" is 0 then you can infer how the minimum stationary point moves up or down (along the y-axis) using certain values of c.
Then we can see that always using either c>0 or c<0 guarantees the minimum stationary point stays in the fourth quadrant (hence the sufficient remark).

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Example below is graph: .

Does the minimum stationary point move outside the fourth quadrant if I use any value of c>0 or any value of c<0?
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