### Video Transcript

Find the equation of the tangent to the curve π¦ is equal to five π₯ cubed plus eight π₯ squared plus five π₯ plus six at the point with π₯-coordinate zero.

The question is asking us to find the equation of the tangent line to our cubic curve. And weβre specifically looking for the tangent to the curve at the point whose π₯-coordinate is zero. Letβs recall what it means for a line to be tangent to our curve π¦ is equal to π of π₯ at a point on our curve. First, letβs write the point of our curve down. Letβs call it π₯ one, π of π₯ one. Our tangent line must have the same slope as our curve does at this point. Our tangent line must also pass through this point. We can use this to help us find the equation of our tangent line.

Remember, the general equation for a straight line which passes through the point π₯ one, π¦ one and has a slope of π is given by π¦ minus π¦ one is equal to π times π₯ minus π₯ one. We know that our tangent line will pass through the point π₯ one, π of π₯ one. And the question tells us weβre looking for the tangent to the curve at the point with π₯-coordinate zero. So π₯ one is equal to zero. So we can find the π¦-coordinate of our curve when π₯ is equal to zero by substituting π₯ is equal to zero into the equation of our curve. And our tangent line must pass through this point. So this means that π¦ one is equal to five times zero cubed plus eight times zero squared plus five times zero plus six.

And we can simplify this expression: itβs equal to six. So our curve passes through the point zero, six. And weβre looking for the tangent to our curve at this point. And remember, our tangent will have the same slope as the curve at this point. So we want to find the slope of our curve at the point zero, six. And remember, we can find the slope of a curve by calculating dπ¦ by dπ₯. And this is equal to the derivative of five π₯ cubed plus eight π₯ squared plus five π₯ plus six with respect to π₯. And we can see that this is the derivative of a polynomial. So we can do this by using the power rule for differentiation.

To differentiate each term, we multiply by the exponent of π₯ and reduce this exponent by one. This gives us 15π₯ squared plus 16π₯ plus five. But we want the slope of our curve at the point zero, six. So we need to substitute π₯ is equal to zero. Substituting π₯ is equal to zero, we get 15 times zero squared plus 16 times zero plus five, which we can calculate to give us five. So the slope of our curve at the point zero, six is equal to five. But remember, this is also the slope of our tangent line, so weβll set our value of π equal to five.

So weβve now shown our tangent line must pass through the point zero, six and have a slope of five. So we can substitute π₯ one is equal to zero, π¦ one is equal to six, and π is equal to five into our general equation for a straight line. This gives us π¦ minus six is equal to five times π₯ minus zero. And we can then simplify and rearrange this equation to get that π¦ minus five π₯ minus six is equal to zero. Therefore, weβve shown that the tangent to the curve π¦ is equal to five π₯ cubed plus eight π₯ squared plus five π₯ plus six at the point with π₯-coordinate zero has the equation π¦ minus five π₯ minus six is equal to zero.