Question

2)
Find the slope of the normal to curve y=x²+3x
at (-2,0)​

Answer 1

Step-by-step explanation:

if $\\ y=x^2+3x$

$\frac{d}{dy}(x^2+3x)=2x+3$

$\\ \frac{dx}{dy} = 2x+3$

the slope of the tangent at (-2,0) is

$\frac{dx}{dy}|(-2,0)=2(-2)+3=-1$

the slope of the normal at the same point is -1

∵ tangent is perpendicular to normal