Question

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

Answer 1

Answer:

$\frac{1}{5}$

Step-by-step explanation:

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

Differentiate with respect to $x$

$\frac{dy}{dx} =3(2x)-5=6x-5$

( Use $\frac{d}{dx}(x^n)=nx^{n-1}$ )

Slope of normal curve is given by $\frac{-1}{\frac{dy}{dx} }$

Slope of normal curve = $\frac{-1}{6x-5}$

To find slope of normal curve at point $(0,0)$ ,put $x=0$ in $\frac{-1}{6x-5}$

Slope of normal curve at point $(0,0)$ = $\frac{-1}{6(0)-5} =\frac{-1}{-5}=\frac{1}{5}$