Video: Finding the π‘₯-Coordinate of the Point on the Curve of a Quadratic Function Where the Tangent Line Is Parallel to the π‘₯-Axis

What is the π‘₯-coordinate of the point where the tangent line to 𝑦 = π‘₯Β² + 12π‘₯ + 11 is parallel to the π‘₯-axis?

03:33

Video Transcript

What is the π‘₯ coordinate of the point where the tangent line to 𝑦 equals π‘₯ squared plus 12π‘₯ plus 11 is parallel to the π‘₯-axis?

So where do we start with this problem? Well we start by looking at the tangent line. As you can see from my sketch, a tangent line is a line that touches a curve. And a key thing about it is that the slope of a tangent at a point is equal to the slope of a curve at that same point. So actually as you can see, it’s slope that we want to look at first.

Well in order for us to find the slope of our curve at any point, we first need to differentiate it. So when we differentiate 𝑦 equals π‘₯ squared plus 12 π‘₯ plus 11, our first term will be to two π‘₯. And just as a reminder of how we’ve done that, we can say that the way we got two π‘₯ was we multiplied the exponent by the coefficient of the π‘₯ term so two multiplied by one. And then it was π‘₯ to power of, and then it’s two minus one because you subtract one from the power.

Okay, great! So first term is two π‘₯. And then our second term is plus 12. And we get that because if we differentiate positive 12π‘₯, you just get 12. And if we differentiate positive 11, you get zero. So great! So we can say that 𝑑𝑦 𝑑π‘₯ is equal two π‘₯ plus 12. So now we want to actually find what the slope is going to be. Well actually if we look at the question, we can see that we want to find the point where the tangent line two 𝑦 plus π‘₯ squared plus 12π‘₯ plus 11 is parallel to the π‘₯-axis.

Now if we think about what that means, if we’ve got a line is parallel to the π‘₯-axis, the slope is going to be equal to the change in 𝑦 over the change in π‘₯. Well the change in 𝑦 is just going to be zero because we have a horizontal line. And then I’ve put here over 𝑛 cause it doesn’t matter what our change in π‘₯ is. We don’t matter which two points we’re going to choose.

Because if we have zero divided by anything, then our answer is going to be zero. So we can say this slope is going to be zero. Okay, great! Now that we know that, we can actually substitute that back into our differential. So we know that 𝑑𝑦 𝑑π‘₯ is equal to zero. So we have the equation zero is equal two π‘₯ plus 12.

So then we subtract 12 from each side and we get negative 12 equals two π‘₯. And then if we divide by two, we get that negative six is equal to π‘₯. So therefore, we can actually say that the π‘₯-coordinate of the point where the tangent line to 𝑦 equals π‘₯ squared plus 12 π‘₯ plus 11 is parallel to the π‘₯-axis is equal to negative six.

Just a quick recap of how we found that, first of all we actually differentiated our function. And this gave us the slope function. Then we actually equated this to zero because we were looking for the point where the actual slope was equal to zero because it was parallel to the π‘₯-axis. We then solved to find π‘₯. And as a tangent actually touches our curve at a point, then therefore if we found the π‘₯-coordinate of negative six for our curve, that’s going to be the same as the π‘₯-coordinate on our tangent line.

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