### Video Transcript

Find the equation of the tangent to the curve π of π₯ equals π₯ squared at its point of intersection with the curve π of π₯ equals 125 over π₯.

So weβve been asked to find the equation of the tangent to this curve π of π₯ at its point of intersection with another curve π of π₯. Letβs begin by determining where these two curves intersect. For the two curves to intersect, the values of the two functions must be the same. We can determine where this occurs by setting π of π₯ equal to π of π₯ and then solving the resulting equation to determine the π₯-coordinate of the point of intersection. Setting π of π₯ equal to π of π₯ gives the equation π₯ squared equals 125 over π₯.

To solve this equation, we first need to multiply both sides by π₯. On the left-hand side, π₯ squared multiplied by π₯ gives π₯ cubed. And on the right-hand side, 125 over π₯ multiplied by π₯ just leaves 125. If π₯ cubed is equal to 125, then π₯ is equal to the cubed root of 125, which is five. So we found the π₯-coordinate at the point of intersection of these two curves. And now we need to find the π¦-coordinate, which will be the value of the functions themselves. We can work this out by substituting π₯ equals five into the equation of either curve because, remember, they must have the same value at this point.

Substituting π₯ equals five into the function π of π₯, we have π of five is equal to five squared, which is 25. And if we want, we can just confirm that π of π₯ has the same value, π of five is equal to 125 over five, which is also equal to 25. This tells us then that the point of intersection of the two curves π of π₯ and π of π₯ is the point five, 25. Now we are asked to find the equation of the tangent to the curve π of π₯ at this point. There are various different methods for finding the equation of a tangent, one of which is using the pointβslope form of the equation of a straight line: π¦ minus π¦ one equals π π₯ minus π₯ one, where π₯ one, π¦ one represent the coordinates of any point on the line and π represents its slope.

Weβve just worked out the coordinates of a point that lies on this line. π₯ one, π¦ one is point five, 25. But we donβt yet know the slope of the line. Remember, though, that the slope of a tangent to a curve is the same as the slope of the curve itself at that point. And we can find the slope at any point on a curve using differentiation. The slope function or first derivative of the function π of π₯, which is given by π prime of π₯, is found by differentiating its equation. π of π₯ is equal to π₯ squared. And we recall that in order to differentiate a polynomial term such as this, we first multiply by the power, thatβs two, and then we reduce the power of π₯ by one, so we have π prime of π₯ is equal to two multiplied by π₯ to the first power, although of course π₯ to the first power is simply π₯.

The general slope function π prime of π₯ is equal to two π₯, but we need to evaluate it specifically at this point. Substituting the π₯-coordinate at the point of intersection then, we have that π prime of five is equal to two multiplied by five, which is equal to 10. As we now know the slope of this tangent and the coordinates of a point that passes through it, we can find its equation.

Substituting the point five, 25 for π₯ one, π¦ one and 10 for the slope π, we have π¦ minus 25 equals 10 multiplied by π₯ minus five. Distributing the parentheses on the right-hand side gives 10π₯ minus 50. And then if we choose to group all of the terms on the left-hand side of the equation, which we can achieve by subtracting 10π₯ and then adding 50, this gives the equation of the tangent.

The equation of the tangent to the curve π of π₯ equals π₯ squared at the point where it intersects the curve π of π₯ equals 125 over π₯ is π¦ minus 10π₯ plus 25 is equal to zero.