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Find the equations of

a)The tangent
b)The normal
to the curve y=2x^3 - x^2 + 4x - 7 at the point where x = 1.

User MattSkala
by
8.4k points

1 Answer

5 votes

Given, equation of curve is


y = ((x - 7))/((x - 2)(x - 3)) .....(1)\\

To find the intersection of given curve with X-axis, Put y = 0 in equation 1, we get,


0 = ((x - 7))/((x - 2)(x - 3)) ......(2) \\

x − 7 = 0 ⟹ x = 7

Thus, the curve cut the X-axis at (7,0).

Now, on differentiating equation of curve w.r.t. x, we get,


(dy)/(dx) \\ = \frac{(x - 2)(x - 3).1 - (x - 7)[(x - 2).1 + (x - 3).1]}{[(x-2)(x - 3) {}^(2)]}

Next answer is in pic.

Find the equations of a)The tangent b)The normal to the curve y=2x^3 - x^2 + 4x - 7 at-example-1
User Eshe
by
8.2k points

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