# Question Video: Using the Maclaurin Series to Find the Equation of a Tangent Mathematics • Higher Education

If the Maclaurin series of the function π is π(π₯) = 2 β (1/6 π₯) + (5/24 π₯Β²) β (7/60 π₯Β³) + (3/80 π₯β΄) + βββ, find the equation of the tangent to the curve of π at π₯ = 0.

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### Video Transcript

If the Maclaurin series of the function π is π of π₯ is equal to two minus one over six multiplied by π₯ plus five divided by 24 multiplied by π₯ squared minus seven divided by 60 multiplied by π₯ cubed plus three divided by 80 multiplied by π₯ to the fourth power and so on. Find the equation of the tangent to the curve of π at π₯ is equal to zero.

The question gives us the Maclaurin series for a function π and asks us to find the equation of the tangent to the curve of π when π₯ is equal to zero. We recall that a straight line will have equation π¦ is equal to ππ₯ plus π where π is the slope of our line and π is the π¦-intercept. We also recall that our function π prime of π₯ gives us the slope of the tangent to the curve of π at π₯. And since the question asks us to find the tangent when π₯ is equal to zero, we can conclude that the slope of our tangent is given by π prime of zero.

Now, we recall that the Maclaurin series of a function is equal to the Taylor series expansion about zero. Namely, this gives us that π of π₯ is equal to π evaluated at zero plus the first derivative of π evaluated at zero multiplied by π₯ plus the second derivative of π evaluated at zero divided by two factorial multiplied by π₯ squared plus the third derivative of π evaluated at zero divided by three factorial multiplied by π₯ cubed. And if we keep going and going, then we get the Maclaurin series for our function π.

From the question weβre given the Maclaurin series of our function π is equal to two plus negative one-sixth multiplied by π₯ plus five over 24 multiplied by π₯ squared plus negative seven divided by 60 multiplied by π₯ cubed. And this keeps going and going. We can see from the definition of our Maclaurin series that the coefficient of π₯ is equal to π prime of zero. Therefore, we must have that the first derivative of π evaluated at zero is equal to negative one-sixth. So we have the slope of our tangent is negative one-sixth giving us the equation π¦ is equal to negative one-sixth multiplied by π₯ plus some constant π.

One way to find the value of π is to find the π₯- and π¦-coordinates of a point on our tangent line. And one way to do this is to notice that since weβre finding the tangent line to the curve of π when π₯ is equal to zero, we must have that the tangent line and the curve of π meet when π₯ is equal to zero. Therefore, we must have that when π₯ is equal to zero, our π¦-value is equal to the function π evaluated at zero. Now, we just need to find the value of the function π evaluated at zero. And we can do this by looking on our Maclaurin series. We can see that the first term in our series is equal to the function π evaluated at zero. So we must have that π zero is equal to two.

So what we have just shown is that the 0.2 is on our tangent line. Therefore, if we substitute the value of π₯ equals zero and the value of π¦ equals two into the equation for our line, we get two is equal to negative one-sixth multiplied by zero plus π. And since negative one-sixth multiplied by zero is just equal to zero, we must have that π is equal to two. So what weβve shown is that for the Maclaurin series π of π₯ given to us in the question, the equation of the tangent to the curve π at π₯ is equal to zero is π¦ is equal to negative one-sixth π₯ plus two.