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I was practicing some maths from the UKMT website's mentoring scheme sheets and I found this question on the sample for sheet 9:

On the same axes, sketch the graphs

$y=(x+1)^3-(x+1)$

$y^2=(x+1)^3-(x+1)$

I also have to identify the points where the graphs intersect.

I found that $(x+1)^3-(x+1) = x(x+1)(x+2)$

From there I found three intersection points: $(0,0), (-1,0)$ and $(-2,0)$

I need to solve $x(x+1)(x+2) = 1$ for another intersection point, but I cant think of any ways to solve it. I was also had no idea on how to sketch a graph for $y^2 = (x+1)^3 - (x+1)$

Chris Daniel
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Can it be solved easily? $$(x+1)^3-(x+1)-1=x^3+3x^2+2x-1$$ The cubic has no rational roots, so there is no simple expression for the $x$-coordinate of the fourth intersection (although some exact expression can be written down). The $x$-coordinate is about $0.324717957\dots$, but the $y$-coordinate is just $1$.

Parcly Taxel
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  • The pdf has no answers attached so I guess It was probably asking for an aproximation as it was a non calculator paper. Thank you for answering. – Chris Daniel Oct 08 '22 at 18:23
  • @ChrisDaniel You have enough reputation to upvote my answer. Please do that here – it helps all of us. – Parcly Taxel Oct 08 '22 at 18:32