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Please help!!!

IMG_20171025_171109.jpgIve done part a and got f(x)=x^2+4x-4+(2x-5)/(x^2-3x+3)
Original post by Ola2898
Ive done part a and got f(x)=x^2+4x-4+(2x-5)/(x^2-3x+3)


Find what f′(1)f'(1) is then use the negative reciprocal of it with the point where x=1x=1 to construct the line.
Original post by Ola2898
IMG_20171025_171109.jpgIve 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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