What is the equation of the line normal to $f (x) = 9 x^{2} - 28 x - 34$ $f(x)=9x^2 - 28x - 34$ at $x = - 1$ $x=-1$ ?

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Answer 1 · 2016-12-29T15:54:25

$46 y - x - 139 = 0$ $46y-x-139=0$

let $y = f (x)$ $" "y=f(x)$

1) find the gradient of the normal

This is done in two parts

(i) find gradient of tangent ${(\frac{d y}{d x})}_{x = - 1}$ $((dy)/(dx))_(x=-1)$

$y = 9 x^{2} - 28 x - 34 \Rightarrow \frac{d y}{d x} = 18 x - 28$ $y=9x^2-28x-34=>(dy)/(dx)=18x-28$

${(\frac{d y}{d x})}_{x = - 1} = 18 \times - 1 - 28 = - 18 - 28 = - 46$ $((dy)/(dx))_(x=-1)=18xx-1-28=-18-28=-46$

(ii) the normal is perpendicular to the tangent .

$m_{t} \times m_{n} = - 1$ $m_txxm_n=-1$

$- 46 \times m_{n} = - 1 \Rightarrow m_{n} = \frac{1}{46}$ $-46xxm_n=-1=>m_n=1/46$

2) find the eqn. of normal

use $y - y_{1} = m_{n} (x - x_{1})$ $" "y-y_1=m_n(x-x_1)$

need to find $y_{1}$ $" " y_1" "$ first

$y (- 1) = 9 \times {(- 1)}^{2} - 28 \times (- 1) - 34$ $y(-1)=9xx(-1)^2-28xx(-1)-34$

$y (- 1) = 9 + 28 - 34 = 3$ $y(-1)=9+28-34=3$

$(x_{1}, y_{1}) = (- 1, 3)$ $(x_1,y_1)=(-1,3)$

$:.y-3=1/46(x--1)$

$46y-138=x+1$

$46y-x-139=0$

What is the equation of the line normal to f(x)=9x^2 - 28x - 34 f(x)=9x2−28x−34f(x)=9x^2 - 28x - 34 at x=-1x=−1x=-1?