Answer
a) $y=-\frac{9}{4}x+\frac{1}{4}$
b) $(-2,2);(-2,-2)$
c) See image.
Work Step by Step
a)
$y^2=x^3+3x^2\\
2yy'=3x^2+6x\\
y'=\frac{3x^2+6x}{2y}$
Plug in $(1,-2)$ into $y′$ to find the gradient
$y'=\frac{3(1)^2+6(1)}{2(-2)}\\
y'=-\frac{9}{4}$
Equation of the tangent line at $(1,-2)$
$y-(-2)=-\frac{9}{4}(x-1)\\
y=-\frac{9}{4}x+\frac{1}{4}$
b) A curve has horizontal tangents when $y'=0$
$0=\frac{3x^2+6x}{2y}\\
0=3x^2+6x\\
3x(x+2)=0\\
x=-2, y=2 \Longrightarrow (-2,2)\\
x=-2, y=-2 \Longrightarrow (-2,-2)\\
x=0, y=0 \Longrightarrow (0,0)$
But at (0,0) the derivative ($y'$) is undefined, not zero.
c) See image.
