Question

The slope of the normal curve y=3x^2-5x is at the point (0,0)

Answer 1

`(1)/(5)`

Explanation:

Consider the curve
`y=3x^2-5x`

Differentiate with respect to
`x`

`(dy)/(dx) =3(2x)-5=6x-5`

( Use
`(d)/(dx)(x^n)=nx^(n-1)` )

Slope of normal curve is given by
`(-1)/((dy)/(dx) )`

Slope of normal curve =
`(-1)/(6x-5)`

To find slope of normal curve at point
`(0,0)` ,put
`x=0` in
`(-1)/(6x-5)`

Slope of normal curve at point
`(0,0)` =
`(-1)/(6(0)-5) =(-1)/(-5)=(1)/(5)`