# Video: Finding the Equation of a Line Normal to a Cubic Curve

What is the equation of the line normal to the curve 𝑦 = (1/3) 𝑥³ − 4𝑥² + 3𝑥 − 4 at 𝑥 = 0?

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### Video Transcript

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

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

Recall that if 𝑚 T is the gradient of the tangent and 𝑚 N is the gradient of the normal, then 𝑚 T multiplied by 𝑚 N equals negative one. In other words, the gradient of the normal is equal to negative one over the gradient of the tangent. Let’s use this information to answer the question.

We are asked to find the equation of a particular line. So, let’s recall the general equation of a line. The general equation of a line 𝑙 is 𝑦 minus 𝑦 one equals 𝑚 times 𝑥 minus 𝑥 one, where 𝑥 one, 𝑦 one is a point that lies on 𝑙 and 𝑚 is the gradient of 𝑙. So, the equation of the line normal to the given curve at the given point is 𝑦 minus 𝑦 one equals 𝑚 N multiplied by 𝑥 minus 𝑥 one, where 𝑥 one, 𝑦 one is a point that lies on the normal and 𝑚 N is the gradient of the normal.

We want to find the equation of the line normal to the given curve at the point 𝑥 equals zero. So by definition, the line passes through the point whose 𝑥-coordinate is equal to zero and 𝑦-coordinate is equal to the curve evaluated at 𝑥 equals zero. Substituting 𝑥 equals zero into the equation for the given curve gives us a 𝑦-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 𝑦 plus four equals 𝑚 N times by 𝑥, where 𝑚 N is the gradient of the normal line.

Let’s 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 𝑥 equals zero is given by d𝑦 by d𝑥, evaluated at 𝑥 equals zero.

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

Recall that the derivative of a constant is zero, and so the constant term, negative four, in 𝑦 differentiates to zero. Simplifying, we obtain that d𝑦 by d𝑥 equals 𝑥 squared minus eight 𝑥 plus three. Evaluating d𝑦 by d𝑥 at 𝑥 equals zero, we obtain the value three. And so, the gradient of the tangent to the curve at 𝑥 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 𝑦 plus four equals negative one over three multiplied by 𝑥.

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