C1 Past Paper Q

Hi I am a little confused by this question (from June 2013 Paper 1R).

I get that x=-1 and x=2 when y=0 so I got y=(x+1)(x-2)

But when x=0, y=4. I'm not sure how to alter my equation to get this.

Thanks

Hi I am a little confused by this question (from June 2013 Paper 1R).

I get that x=-1 and x=2 when y=0 so I got y=(x+1)(x-2)

But when x=0, y=4. I'm not sure how to alter my equation to get this.

Thanks

I think you're getting a little confused with quadratics.
we have two turning points here, one at x=0 and one at x=2. What do you know about turning points?
you also have the values where y=0. If you expand (x+1)(x-2) do you get a cubic? The one at x=2 looks different - can you use that?

Because the curve TOUCHES the axis at x=2, the root is '.....'?
ie y=(x+1)(x-2)^?
tbh, Im not sure of the relevance of them giving you the y-intercept soon as to aswer part a, you would simply expand out and read off a, b and c. Hope that helps.

No I know that (x+1)(x-2) doesn't make a cubic, I'm not that stupid. I need to add to what I have to get a cubic equation which is correct.

At turning points the gradient goes down to 0, that's all I know, I haven't done C2.

So basically I just square (x-2) because the curves turns at x=2?



Oh I see - so is that a general rule then? Where it touches the x-axis it's repeated and squared.

Yep - now to finish up, you may want to look at my now edited post

Well when y=4 x=0 and when x=0 the other terms with x in will equal 0 so it'll leave y=4 so c must be 4.

Cheers

Where did you get the (x-2)^2 part from? I've never even seen that before in a C1 paper, or the textbook. Neither have any of my class.


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