Question 1023346
Differentiate implicitly,
{{{2ydy=3x^2dx+6xdx}}}
{{{(2y)dy=(3x^2+6x)dx}}}
{{{dy/dx=(3x^2+6x)/(2y)}}}
So at (1,2),
{{{dy/dx=(3(1)^2+6(1))/(2(2))}}}
{{{dy/dx=(3+6)/4}}}
{{{dy/dx=9/4}}}
Using the point-slope form,
{{{y-2=(9/4)(x-1)}}}
{{{y-2=(9/4)x-9/4}}}
{{{y=(9/4)x-9/4+8/4}}}
{{{y=(9/4)x-1/4}}}
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b) When the derivative equals zero, the numerator must equal zero so,
{{{3x^2+6x=0}}}
{{{3x(x+2)=0}}}
Two solutions,
{{{x=0}}}
However, when {{{x=0}}}, {{{y=0}}}, so then the derivative is undefined since there is a division by zero so this {{{x}}} value is not allowed.
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{{{x+2=0}}}
{{{x=-2}}}
So here there are two tangents,
{{{y^2=(-2)^3+3(-2)^2}}}
{{{y^2=-8+12}}}
{{{y^2=4}}}
{{{y=2}}} and {{{y=-2}}}
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*[illustration CS1.JPG].
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c) Fred or it might be George.