Question

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.

Answer 1

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.