Question

a) The curve with equation y2 = x3 + 3x2 is called the Tschirnhausen cubic. Find an equation of the tangent line to this curve at the point (1, 2). y = Correct: Your answer is correct. (b) At what points does this curve have horizontal tangents? (x, y) = -2 Incorrect: Your answer is incorrect. (smaller y-value) (x, y) = 2 Incorrect: Your answer is incorrect. (larger y-value)

Answer 1

Answer:

Tangent at (1,2) has value $\frac{15}{4}$. And (-2, 4) is the point of intersection of horizontal tangent.

Step-by-step explanation:

Given curve equation,

$y^2=x^3+3x^2\hfill (1)$

To find tangent at (x,y)=(1,2) and point of intersection of horizontal tangent, differentiate (1) withrespect to x we get,

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

$\implies \frac{dy}{dx}=\frac{3(x^2+2x)}{2y}$

At (1, 2), tangent line is,

$\frac{dy}{dx}|_{(1,2)}}=\frac{15}{4}$

To find point of intersection of horizontal tangent we have to do,

$\frac{dy}{dx}=0$

$\implies x(x+2)=0\implies x=0 or -2$

Thus,

At x=0, y=0

but snce,

$\lim_{y\to 0}\frac{dy}{dx}\to \infty$

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.