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# C1 graph sketching question watch

1. Sketch the curve with equation:
y = (x-1)^3 - 4(x-1)
Showing the coordinates of the points where the curve crosses the x-axis

_____

Here's my working:
(x-1)(x-1)(x-1) - 4(x-1)
(x-1)(x^2-2x+1) - 4x+4
x^3 - 2x^2 + x - x^2 + 2x - 1 - 4x + 4
x^3 - 3x^2 +3x - 1 - 4x + 4
x^3 - 3x^2 - x + 3

I don't know what to do after i've worked out x^3 - 3x^2 - x + (unless i've done it wrong).
2. Factorise that into 3 brackets to get all the intercepts.
As its +x^3 its drawn from the top right.
3. ah don't think i've been taught that, I know how to factorise it into a cubic if the +3 wasn't there.
4. (Original post by blankboi)
ah don't think i've been taught that, I know how to factorise it into a cubic if the +3 wasn't there.
can you get it into a quadratic and a single bracket.
The number in the single bracket would have to multiply with C in the quadratic to give +3.
If that makes any sense.
5. (Original post by blankboi)
Sketch the curve with equation:
y = (x-1)^3 - 4(x-1)
Showing the coordinates of the points where the curve crosses the x-axis

_____

Here's my working:
(x-1)(x-1)(x-1) - 4(x-1)
(x-1)(x^2-2x+1) - 4x+4
x^3 - 2x^2 + x - x^2 + 2x - 1 - 4x + 4
x^3 - 3x^2 +3x - 1 - 4x + 4
x^3 - 3x^2 - x + 3

I don't know what to do after i've worked out x^3 - 3x^2 - x + (unless i've done it wrong).
There's a really easy way to factorise this.

Observe that

If you now expand the contents of the square bracket, you'll get a quadratic that factorises

That will leave you with all 3 points at which it crossss the x-axis

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