# Video: APCALC03AB-P1A-Q24-319125857816

What is the number of vertical tangents to the graph of 𝑦² = 𝑥² − 3𝑥 + 2?

04:05

### Video Transcript

What is the number of vertical tangents to the graph of 𝑦 squared equals 𝑥 squared minus three 𝑥 plus two?

First, let′s recall what we know about vertical tangents. If the tangent to a graph at a given point is vertical, then that means that the slope of the line and hence the slope of the curve itself at that point is infinite. The slope of a curve is given by its first derivative d𝑦 by d𝑥. So we now know we′re looking for the number of points on this graph, where its slope d𝑦 by d𝑥 is infinite. We′re therefore going to need to find an expression for d𝑦 by d𝑥. And your first thought may be that we need to rearrange this equation slightly to make 𝑦 the subject. By square rooting each side, we get that 𝑦 is equal to plus or minus the square root of 𝑥 squared minus three 𝑥 plus two. But that plus or minus will cause a problem for us when we′re looking to differentiate.

Instead, we′re going to apply the method of implicit differentiation in order to find this derivative. We′re going to differentiate both sides of the equation with respect to 𝑥. Now, the right-hand side is no problem because it′s just a polynomial in 𝑥. And so we can apply the power rule of differentiation in order to find its derivative. The derivative of 𝑥 squared is two 𝑥. The derivative of negative three 𝑥 is negative three. And the derivative of any constant is zero. So we have that the derivative with respect to 𝑥 of the right-hand side of our equation is two 𝑥 minus three. But what about the left-hand side? Here we have a function in 𝑦. It′s 𝑦 squared. But we′re differentiating with respect 𝑥. So this is where we need to recall the method of implicit differentiation.

Implicit differentiation is just an application of the chain rule. It tells us that if 𝑓 is a function of 𝑦 and 𝑦 is a function of 𝑥. Then the derivative of 𝑓 with respect to 𝑥 is equal to the derivative of 𝑓 with respect to 𝑦 multiplied by the derivative of 𝑦 with respect to 𝑥. Here, our function 𝑓 is 𝑦 squared. So let′s apply this method. The derivative of 𝑓 with respect to 𝑦, that′s the derivative of 𝑦 squared with respect to 𝑦, is two 𝑦. And then, we multiply by d𝑦 by d𝑥. So we have that two 𝑦 d𝑦 by d𝑥 is equal to two 𝑥 minus three. We can rearrange this equation to make d𝑦 by d𝑥 the subject. And thus, we have an expression for our first derivative d𝑦 by d𝑥 in terms of both 𝑥 and 𝑦. d𝑦 by d𝑥 is equal to two 𝑥 minus three over two 𝑦.

Now remember, we′re looking to determine at how many points on this graph the tangents to the curve are vertical. And we′ve already said that this will be the case when d𝑦 by d𝑥 is infinite. Well, looking at our expression for d𝑦 by d𝑥, we can see that it will be infinite when the denominator of the fraction is equal to zero. The denominator is two 𝑦. And if two 𝑦 must be equal to zero, it follows that 𝑦 itself must be equal to zero.

We don′t yet know though at how many points on the curve 𝑦 is equal to zero. So we need to find the corresponding 𝑥-values, which we do by substituting 𝑦 equals zero back into the equation of the original curve. Doing so gives zero equals 𝑥 squared minus three 𝑥 plus two. This quadratic can be factored. It′s equal to 𝑥 minus two multiplied by 𝑥 minus one. To solve, we take each factor in turn, set it equal to zero, and solve the resulting linear equation, giving 𝑥 is equal to two or 𝑥 is equal to one. We know then that the tangents to this graph will be vertical at two points, the points one, zero and two, zero.

To answer the question then what is the number of vertical tangents or how many vertical tangents are there to this graph, we say that there are two.

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