### Video Transcript

What is the equation of the line
normal to the curve π¦ equals a third π₯ cubed minus four π₯ squared plus three π₯
minus four at π₯ equals zero?

Before we start answering the
question, letβs quickly recall the definition of the line normal to a curve at a
particular point. Suppose we are given a curve π¦
equals π of π₯ and a point π₯ one, π¦ one that lies on it. Then, the line normal to the curve
π¦ at the point π₯ one, π¦ one is the line passing through the point π₯ one, π¦ one,
which is perpendicular, i.e. 90 degrees, to the tangent to the curve π¦ at the point
π₯ one, π¦ one.

Recall that if π T is the gradient
of the tangent and π N is the gradient of the normal, then π T multiplied by π N
equals negative one. In other words, the gradient of the
normal is equal to negative one over the gradient of the tangent. Letβs use this information to
answer the question.

We are asked to find the equation
of a particular line. So, letβs recall the general
equation of a line. The general equation of a line π
is π¦ minus π¦ one equals π times π₯ minus π₯ one, where π₯ one, π¦ one is a point
that lies on π and π is the gradient of π. So, the equation of the line normal
to the given curve at the given point is π¦ minus π¦ one equals π N multiplied by
π₯ minus π₯ one, where π₯ one, π¦ one is a point that lies on the normal and π N is
the gradient of the normal.

We want to find the equation of the
line normal to the given curve at the point π₯ equals zero. So by definition, the line passes
through the point whose π₯-coordinate is equal to zero and π¦-coordinate is equal to
the curve evaluated at π₯ equals zero. Substituting π₯ equals zero into
the equation for the given curve gives us a π¦-value of negative four. Hence, the point zero, negative
four lies on the normal line in question. Substituting this into the general
equation for the normal line, we obtain π¦ plus four equals π N times by π₯, where
π N is the gradient of the normal line.

Letβs now find the gradient of the
normal line. We know that the gradient of the
line normal to the given curve at the given point is equal to negative one over the
gradient of the tangent to the given curve at the given point. The gradient of the tangent to the
given curve at π₯ equals zero is given by dπ¦ by dπ₯, evaluated at π₯ equals
zero.

Since the curve π¦ is a sum of
terms of the form ππ₯ to the power of π, where π and π are constants. And since the derivative of a sum
of terms is the sum of the derivatives of the terms. We will find dπ¦ by dπ₯ by
differentiating each term in π¦ using the formula where we multiply the coefficient
π by the exponent π and reduce the exponent by one. Doing so, we obtain that dπ¦ by dπ₯
equals a third times by three times by π₯ squared minus four times two times by π₯
to the power of one plus three times one times by π₯ to the power of zero plus
zero.

Recall that the derivative of a
constant is zero, and so the constant term, negative four, in π¦ differentiates to
zero. Simplifying, we obtain that dπ¦ by
dπ₯ equals π₯ squared minus eight π₯ plus three. Evaluating dπ¦ by dπ₯ at π₯ equals
zero, we obtain the value three. And so, the gradient of the tangent
to the curve at π₯ equals zero is equal to three. Substituting this into our equation
for the normal line, we obtain that the equation for the normal line is π¦ plus four
equals negative one over three multiplied by π₯.

Rearranging this equation by
subtracting four from each side, we obtain π¦ equals negative one over three
multiplied by π₯ minus four. So, our final answer is that the
equation of the line normal to the given curve at the point π₯ equals zero is π¦
equals negative one over three multiplied by π₯ minus four.