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Find the equation of the normal to the curve y=3x^2-2x-1 which is parallel to the line y=x-3 .

Hi I'm having trouble figuring out when to use which gradients as initially the gradient you get is one which you hen convert into -1 and then finally use the gradient of one again.

The answer is y=x-17/12 or 12y=12x-17 (same thing)

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A normal at some point is perpendicular to the tangent line at that point. The tangent line's slope at any point is

$$y'=6x-2\;\;\text{and we want this to be perpendicular to a line with slope}\;\;1\implies$$

$$6x-2=-1\implies x=\frac16$$

and since

$$y\left(\frac16\right)=3\frac1{36}-2\frac16-1=\frac1{12}-\frac13-1=-\frac54\implies \left(\frac16\,,\,-\frac54\right)$$

is the tangency point , so the normal there is

$$y+\frac54=x-\frac16\implies y=x-\frac{17}{12}$$

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