Question Video: Finding the Equation of the Normal to the Curve of a Polynomial Function at a Point with the Given π‘₯-Coordinate Mathematics • Higher Education

Find the equation of the normal to the curve 𝑦 = βˆ’2π‘₯Β³ βˆ’ 7π‘₯Β² + 2 at π‘₯ = βˆ’2.

04:50

Video Transcript

Find the equation of the normal to the curve 𝑦 is equal to negative two π‘₯ cubed minus seven π‘₯ squared plus two at π‘₯ is equal to negative two.

We’re given a curve which is defined by a cubic polynomial. And we need to determine the equation of the normal to this curve at the point where π‘₯ is equal to negative two. To do this, we’re first going to need to recall what we mean by the normal of a curve at a point and how to find its equation. We’ll start by recalling the equation of a straight line. It’s given by 𝑦 minus 𝑦 one is equal to π‘š times π‘₯ minus π‘₯ one, where π‘š is the slope of our straight line and our straight line passes through the point π‘₯ one, 𝑦 one. The reason we need this is the normal to a curve at a point is a straight line which is perpendicular to our tangent line at that point.

Therefore, to find the equation of the normal to the curve at this point, we just need to find its slope and the point which it passes through. Let’s start by finding a point that our normal line must pass through. In the question, we were told this is the normal line to the curve at the point on the curve where π‘₯ is equal to negative two. Therefore, the normal line must pass through the point on our curve where π‘₯ is equal to negative two. Of course, the π‘₯-coordinate of this point will be negative two. And we can find the 𝑦-coordinate by substituting negative two into the equation for our curve. We get 𝑦 one will be equal to negative two multiplied by negative two cubed minus seven times negative two squared plus two. We can then just calculate this expression. We get 𝑦 one is 16 minus 28 plus two, which we can calculate is equal to negative 10.

So we found values for 𝑦 one and π‘₯ one. All we need to do now is find the value for our slope π‘š. To do this, remember the normal line to our curve at a point will be perpendicular to the tangent line to the curve at this point. So let’s say that our curve is 𝑦 is equal to some function 𝑓 of π‘₯. In this case, 𝑓 of π‘₯ will be negative two π‘₯ cubed minus seven π‘₯ squared plus two. We know how to find the slope of the tangent line at this point. The tangent line will have the same slope as our curve at this point. In other words, it will be 𝑓 prime of negative two. But the normal line needs to be perpendicular to this line. In other words, π‘š will be equal to negative one divided by 𝑓 prime of negative two.

And there’s one small thing worth pointing out here. This won’t make sense if 𝑓 prime of negative two is equal to zero. However, if this is the case, we can do something slightly differently. If 𝑓 prime of negative two is equal to zero, this means that our tangent line is a horizontal line. And for our normal line to be perpendicular to a horizontal line, our normal line would just be a vertical line. So if this were the case, we would need to find an equation in the form π‘₯ is equal to some constant π‘Ž. So let’s try and find the slope of our normal line π‘š. We see we need to find 𝑓 prime of negative two. This means we’re going to need to first find an expression for 𝑓 prime of π‘₯.

To find 𝑓 prime of π‘₯, we need to recall this is the derivative of 𝑓 of π‘₯ with respect to π‘₯. That’s the derivative of negative two π‘₯ cubed minus seven π‘₯ squared plus two with respect to π‘₯. In this case, we’re just evaluating the derivative of a polynomial. We can do this term by term by using the power rule for differentiation. We want to multiply by our exponent of π‘₯ and then reduce this exponent by one. We get 𝑓 prime of π‘₯ is negative six π‘₯ squared minus 14π‘₯.

We’re now ready to find the value of 𝑓 prime of negative two. Remember, this will be the slope of the tangent line to our curve at the point on our curve where π‘₯ is equal to negative two. We substitute π‘₯ is equal to negative two into our expression for 𝑓 prime of π‘₯. We get 𝑓 prime of negative two is equal to negative six times negative two squared minus 14 times negative two. And then we can just calculate this expression. We get 𝑓 prime of negative two is equal to four. We can then substitute this into our expression for π‘š or we can do this by remembering we just need to take the negative reciprocal of this value. Either way, we get π‘š is equal to negative one-quarter.

Now that we found a point on our normal line and its slope, we just need to substitute these values into our equation for a line. Substituting π‘₯ one is equal to negative two, 𝑦 one is equal to negative 10, and π‘š is equal to negative one-quarter into our equation for a straight line, we get 𝑦 minus negative 10 is equal to negative one-quarter multiplied by π‘₯ minus negative two. And we can simplify this. First, 𝑦 minus negative 10 is the same as 𝑦 plus 10. Similarly, π‘₯ minus negative two is the same as π‘₯ plus two. So we were able to simplify this to get 𝑦 plus 10 is equal to negative one-quarter multiplied by π‘₯ plus two.

And we could simplify this by distributing negative one-quarter over our parentheses. However, instead, we’re going to multiply both sides of our equation through by four. This gives us four 𝑦 plus 40 is equal to negative one multiplied by π‘₯ plus two. Once again, we could distribute the negative over our parentheses. However, we’re just going to add π‘₯ plus two to both sides of the equation. This gives us four 𝑦 plus 40 plus π‘₯ plus two is equal to zero. Finally, we can simplify and rearrange this equation to get four 𝑦 plus π‘₯ plus 42 is equal to zero, which is our final answer.

Therefore, we were able to show the equation of the normal to the curve 𝑦 is equal to negative two π‘₯ cubed minus seven π‘₯ squared plus two at the point where π‘₯ is equal to negative two is given by the equation four 𝑦 plus π‘₯ plus 42 is equal to zero.

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