Question

How do you find the equation tangent to `y=x^4-3x^2+2` at Point: (1,0)?

Answer 1

`2x+y-2=0`.

Explanation:

We recall that, slope of the tgt. to the given curve at the pt. `(1,0)` is `[dy/dx]_(x=1,y=0)`.

`y=x^4-3x^2+2 rArr dy/dx=4x^3-6x`

`rArr [dy/dx]_(x=1,y=0)=4-6=-2`.

Thus, the slope of the tgt. is `-2`, it passes thro. the pt. `(1,0)`.

Hence, eqn. of the tgt. is`: y-0=-2(x-1), i.e., 2x+y-2=0`.