What is the equation of the line tangent to $f(x)=(x-1)^2(2x+4)$ at $x=0$ ?

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Answer 1 · 2017-12-04T01:03:48

$y=-6x+4$

Solve for $y$ by plugging $x=0$ into the given equation.

$f(0)=(-1)^2(4)=4$

Find the slope of the tangent line by solving for the second derivative. Use the product rule.

$f'(x)=2(x-1)^2+(2x+4)2(x-1)$
$f'(x)=2(x-1)^2+(2x+4)(2x-2)$

Plug in $x=0$ to find the slope of the tangent line at (0, 4) point.

$f'(0)=2(-1)^2+(4)(-2)$
$f'(0)=2-8=-6$

With your knowledge of the slope and (x,y) coordinate, use the following formula to find the equation of the tangent line.

$y-y_1=m(x-x_1)$
$y-4=-6(x-0)$
$y=-6x+4$

What is the equation of the line tangent to f(x)=(x-1)^2(2x+4) f(x)=(x-1)^2(2x+4) at x=0 x=0 ?