Question

(I) find the equation of the tangent to the curve y=x^2-2x-3 at the point (3,0)? (II) A curve is such that day/dx=15x^2-12x.Given that it passes through (1,3). Find its equation?

Answer 1

(I) `y=7x-21` and (II) `y=5x^3-6x^2+4`

Explanation:

(I) The slope of the tangent to a curve `f(x)` at `(x_0,f(x_0))` is given by `f'(x_0)` i.e. the value of derivative of `f(x)` at `x_0`.

The function is `y=x^2-2x-3` - and observe that at `x=3`, `y=3^2-2xx3-3=0`, hence as `(dy)/(dx)=2x-2`,

the slope of tangent is `2xx3-2=7` andas it passes through `(3,0)`, its equation is

`y-0=7(x-3)` or `y=7x-21`

graph{(x^2-2x-3-y)(7x-21-y)=0 [-8.04, 11.96, -4.64, 5.36]}

(II) As `dy/dx=15x^2-12x` we have `y=int(15x^2-12x)dx`

= `5x^3-6x^2+c`

As it passes through `(1,3)`, we have

`3=5*1^3-6*1^2+c` or `3=5-6+c` or `c=4`

Hence function is `y=5x^3-6x^2+4`

graph{y=5x^3-6x^2+4 [-9.04, 10.96, -2.36, 7.64]}