Answer :
Answer:
Tangent at (1,2) has value [tex]\frac{15}{4}[/tex]. And (-2, 4) is the point of intersection of horizontal tangent.
Step-by-step explanation:
Given curve equation,
[tex]y^2=x^3+3x^2\hfill (1)[/tex]
To find tangent at (x,y)=(1,2) and point of intersection of horizontal tangent, differentiate (1) withrespect to x we get,
[tex]2y\frac{dy}{dx}=x^3+3x^2[/tex]
[tex]\implies \frac{dy}{dx}=\frac{3(x^2+2x)}{2y}[/tex]
At (1, 2), tangent line is,
[tex]\frac{dy}{dx}|_{(1,2)}}=\frac{15}{4}[/tex]
To find point of intersection of horizontal tangent we have to do,
[tex]\frac{dy}{dx}=0[/tex]
[tex]\implies x(x+2)=0\implies x=0 or -2[/tex]
Thus,
At x=0, y=0
but snce,
[tex]\lim_{y\to 0}\frac{dy}{dx}\to \infty[/tex]
at (0,0) there exist a vertical tangent. And,
At x=-2, y=4.
Thus (-2, 4) is the point of intersection of horizontal tangent.