Check the solution to "What is the equation of the normal line..."

Ariel Collier

Answered question

2022-04-09

What is the equation of the normal line of

f (x) = e^{x} - x^{3}

at x=0?

Assorrymarf0cgr

Beginner2022-04-10Added 10 answers

Given

f (x) = e^{x} - x^{3}

Then

f^{'} (x) = e^{x} - 3 x^{2}

f (0) = e^{0} - 0 = 1

So the graph of the function passes through (0,1)

f^{'} (0) = e^{0} - 0 = 1

So the slope of the tangent at (0,1) is 1.
If the slope of the tangent is m then the slope of the normal is

- \frac{1}{m}

. So in our example the slope of the normal is

- \frac{1}{1} = - 1

The equation of a line of slope m passing through a point

(x_{0}, y_{0})

can be written as:

(y - y_{0}) = m (x - x_{0})

In our case we have

(x_{0}, y_{0}) = (0, 1)

and m=-1, so:
(y-1)=-1(x-0)=-x
That is y=-x+1 in slope intercept form.

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