### Video Transcript

A curve passes through the point nine, four. The slope of its tangent at a point on the curve π₯, π¦ is given by root π₯ times five π₯ plus three. Find the equation of the tangent at the point when π₯ is equal to one.

To answer this question, we first begin by spotting that weβve been given an expression for the slope of the tangent to the curve. Now, if the equation of a curve is π¦ in terms of π₯, then we find the slope of the tangent at the point on that curve by differentiating π¦ with respect to π₯. So in this case, dπ¦ by dπ₯ is root π₯ times five π₯ plus three.

Now, the question wants us to find the equation of the tangent at a point where π₯ is equal to one. Now, we can find the slope of the tangent when π₯ is equal to one by substituting π₯ is equal to one into our expression for the derivative. Thatβs the square root of one times five times one plus three, which is simply eight.

Now, to find the equation of the tangent, we use the formula for the equation of a straight line. Itβs π¦ minus π¦ one equals π times π₯ minus π₯ one. We have π, we have the slope of our tangent, but weβre not quite sure where it passes through. We know it passes through a point with π₯-coordinate one. But whatβs the π¦-coordinate at that point?

Well, to find that, weβre going to need to find the equation of the curve. And since integration and differentiation are the reverse processes of one another, we find an expression for π¦ by integrating dπ¦ by dπ₯ with respect to π₯. Now, letβs rewrite the square root of π₯ as π₯ to the power of one-half. And we see that π¦ is equal to the indefinite integral of π₯ to the power of one-half times five π₯ plus three with respect to π₯. And the simplest way to evaluate this integral is simply to distribute our parentheses.

When we multiply π₯ to the power of one-half by five π₯, we add the exponents one-half and one. So we get five π₯ to the power of three over two. And then when we multiply π₯ to the power of a half by three, we get three π₯ to the power of one-half. Weβll integrate term by term, remembering that to integrate a power term whose exponent is not equal to negative one, we add one to the exponent and then divide by that new value.

So when we integrate five π₯ to the power of three over two, we get five π₯ to the power of five over two divided by five over two. Similarly, when we integrate three π₯ to the power of one-half, we get three π₯ to the power of three over two divided by three over two. And then, of course, we need a constant of integration πΆ. This simplifies to π¦ equals two π₯ to the power of five over two plus two π₯ to the power of three over two plus πΆ.

Now, to find the equation of our curve, weβll use the fact that we know it passes through the point nine, four. In other words, when π₯ is equal to nine, π¦ is equal to four. Weβre going to substitute these values in. That gives us four equals two times nine to the power of five over two plus two times nine to the power of three over two plus πΆ. This right-hand side becomes 486 plus 54 plus πΆ. And if we subtract 486 and 54 from both sides, we find πΆ is equal to negative 536. And when we replace πΆ with this value, we find that the equation of our curve is π¦ equals two π₯ to the power of five over two plus two π₯ to the power of three over two minus 536.

Remember, weβre looking to find the value of π¦ when π₯ is equal to one. So we let π₯ is equal to one, and we get π¦ equals two times one to the power of five over two plus two times one to the power of three over two minus 536, which is negative 532. We now know the point that our tangent passes through. Itβs when π₯ is equal to one and π¦ is equal to negative 532.

And so, with the equation of our straight line alongside π equals eight, we let π₯ sub one be equal to one and π¦ sub one be equal to negative 532. And π¦ minus negative 532 is therefore equal to eight π₯ minus one. We distribute our parentheses, and then weβre going to subtract eight π₯ and add eight to both sides. And in doing so, we found the equation of the tangent to our curve at the point where π₯ is equal to one. Itβs π¦ minus eight π₯ plus 540 equals zero.