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CORE 1 differentiation question help

The curve with equation y = x^2 + 2x passes through the origin, O.
(i) Find an equation for the normal to the curve at O. [4]

So I did the following for part (i).....

dy/dx= 2x +2
Therefore, negative recip = 1/(2x+2)

The mark scheme says....
= 2x + 2
grad of tangent = 2
grad of normal = -1/2

I get how to differentiate but I don't understand why the gradient is just -1/2, surely the +2 from (2x+2) should feature in the gradient.

Could someone please give some method or something to show me why Im wrong and yeah guidance please :biggrin: thanks
It's just a rule. If m is the gradient of a tangent, -1/m is the gradient of the normal.
Original post by Audi
The curve with equation y = x^2 + 2x passes through the origin, O.
(i) Find an equation for the normal to the curve at O. [4]

So I did the following for part (i).....

dy/dx= 2x +2
Therefore, negative recip = 1/(2x+2)

The mark scheme says....
= 2x + 2
grad of tangent = 2
grad of normal = -1/2

I get how to differentiate but I don't understand why the gradient is just -1/2, surely the +2 from (2x+2) should feature in the gradient.

Could someone please give some method or something to show me why Im wrong and yeah guidance please :biggrin: thanks


But they specifically ask for the equation of the normal at O where x=0. So substituting there gives that.

Instead of finding an expression giving the gradient of the perpendicular its simpler to find the gradient of the tangent at that point and then use the product of gradients rule to find gradient of the normal.
Reply 3
Original post by Audi
The curve with equation y = x^2 + 2x passes through the origin, O.
(i) Find an equation for the normal to the curve at O. [4]

So I did the following for part (i).....

dy/dx= 2x +2
Therefore, negative recip = 1/(2x+2)



You said negative but you forgot negative

So negative reciprocal = 12x+2\dfrac{-1}{2x+2}

Now put x=0 in as you are looking at the Normal at (0,0)

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