What is the equation of the normal line of f(x)=12x3−4x2−5x at x=-2?

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Egerlandsq0z · Accepted Answer

Normal line will be perpendicular to the tangent line. As we know that the product of perpendicular gradients is always -1 we can find the gradient of the normal from the gradient of the tangent.
We compute the slope of the tangent by evaluating the first derivative:

f^{'} (x) = 36 x^{2} - 8 x - 5

f'(-2)=144+16-5=155
So the slope of the normal

(m_{n})

can be found by:

m_{n} \cdot 155 = - 1 \Rightarrow m_{n} = - \frac{1}{155}

To calculate the equation of the normal line we use
y-b=m(x-a)
To use this, we need a point on the line. We know that x = -2 is on the line, so we evaluate the original function at this point to get :
f(-2)=-102 hence:

y - (- 102) = - \frac{1}{155} (x - (- 2))

y + 102 = - \frac{1}{155} x - \frac{2}{155}

y = - \frac{1}{155} x - \frac{15812}{155}

What is the equation of the normal line

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