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

$y = \frac{(x - 7)}{(x - 2)(x - 3)} .....(1)$

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

$0 = \frac{(x - 7)}{(x - 2)(x - 3)} ......(2)$

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,

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

Next answer is in pic.