What is the equation of the line normal to $f (x) = (x - 1) {(x + 2)}^{2}$ $f(x)=(x-1)(x+2)^2$ at $x = 1$ $x=1$ ?

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Answer 1 · 2016-09-05T11:48:00

$y = \frac{1}{9} x + \frac{1}{9}$ $color(blue)(y=1/9x+1/9)$

$determine the gradient of f (x)$ $color(blue)("determine the gradient of "f(x))$

Expanding the brackets

$f (x) \equiv (x - 1) (x^{2} + 4 x + 4)$ $f(x)-=(x-1)(x^2+4x+4)$

$f (x) \equiv x^{3} + 4 x^{2} + 4 x - x^{2} - 4 x - 4$ $f(x)-=x^3+4x^2+4x" "-x^2-4x-4$

$= x^{3} + 3 x^{2} - 4$ $=x^3+3x^2-4$

$f'=3x^2+6x$

$f'(1)=3(1)^2+6(1) = 9$

$"The above value of 9 is the gradient of " f(x)" at "x=1$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$color(blue)("Determine the equation of the straight line")$

So the gradient of the straight line normal to this is $-1/9$ and it will pass through the point $P_1->(x_1,y_1)$ where $x_1=1 larr" given"$

So at $P_1$

$f(1)=y_1=(x_1)^3+3(x_1)^2-4$

$y_1=1+3-4=0$

Giving $y=mx+c" " ->" "y_1=-1/9x_1+c$

$c=0+1/9(1) = +1/9$

$color(blue)(y=1/9x+1/9)$

What is the equation of the line normal to f(x)=(x-1)(x+2)^2 f(x)=(x−1)(x+2)2 f(x)=(x-1)(x+2)^2 at x=1x=1 x=1?