Nevaeh Glass

2022-04-10

What is the equation of the normal line of f(x)=(x-1)(x+4)(x-2) at x=3?

ysnlm8eut

Explanation:
before differentiating , distribute the brackets.
start with $\left(x-1\right)\left(x+4\right)={x}^{2}+3x-4$
$\left({x}^{2}+3x-4\right)\left(x-2\right)={x}^{3}-2{x}^{2}+3{x}^{2}-6x-4x+8$
$f\left(x\right)={x}^{3}+{x}^{2}-10x+8$
The derivative of f(x) is the gradient of the tangent and f'(3)
the value of the tangent.
${f}^{\prime }\left(x\right)=3{x}^{2}+2x-10$
and ${f}^{\prime }\left(3\right)=3{\left(3\right)}^{2}+2\left(3\right)-10=27+6-10=23=m$
For 2 perpendicular lines , the product of their gradients
is -1
let ${m}_{1}=$ gradient of normal
then $m\cdot {m}_{1}=-1\to 23\cdot {m}_{1}=-1⇒{m}_{1}=-\frac{1}{23}$
$f\left(3\right)=\left(2\right)\left(7\right)\left(1\right)=14⇒\left(3,14\right)$ is normal point
equation of normal is $y-b=m\left(x-a\right)m=-\frac{1}{23},\left(a,b\right)=\left(3,14\right)$
so $y-14=-\frac{1}{23}\left(x-3\right)$
(multiply by 23 to eliminate fraction )
$23y-322=-x+3⇒23y+x-325=0$

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