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# Please help!!! watch

1. Ive done part a and got f(x)=x^2+4x-4+(2x-5)/(x^2-3x+3)
2. (Original post by Ola2898)
Ive done part a and got f(x)=x^2+4x-4+(2x-5)/(x^2-3x+3)
Find what is then use the negative reciprocal of it with the point where to construct the line.
3. (Original post by Ola2898)
Ive done part a and got f(x)=x^2+4x-4+(2x-5)/(x^2-3x+3)
Looks fine so far. Are you familiar with finding the normal to a curve at a point?

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