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Question Video: Finding the Equation of a Curve given Its Second Derivative and the Equation of Its Tangent at a Point by Using Integration Mathematics

The second derivative of a function is 6π‘₯ and the equation of the tangent to its graph at the point (βˆ’2, βˆ’4) is 6π‘₯ βˆ’ 𝑦 + 8 = 0. Find the equation of the curve.

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Video Transcript

The second derivative of a function is six π‘₯ and the equation of the tangent to its graph at the point negative two, negative four is six π‘₯ minus 𝑦 plus eight is equal to zero. Find the equation of the curve.

In this question, we’re given the second derivative of a function and an equation for the tangent to its graph at the point negative two, negative four. We need to use this to find the equation of the curve. To do this, let’s start by saying the equation of the curve is in the form 𝑦 is equal to 𝑓 of π‘₯. This then allows us to represent some of the information given to us in the question. For example, the second derivative of this function is six π‘₯, so 𝑓 double prime of π‘₯ is equal to six π‘₯. And we can actually use this to find an expression for 𝑓 of π‘₯.

To do this, remember, we’re taking the second derivative of 𝑓 of π‘₯ with respect to π‘₯. This means when we differentiate 𝑓 of π‘₯ twice with respect to π‘₯, we get six π‘₯. 𝑓 of π‘₯ is an antiderivative of the antiderivative of six π‘₯. And we can recall that the indefinite integral of a function gives us its most general antiderivative. Therefore, we can find the indefinite integral of 𝑓 double prime of π‘₯ twice with respect to π‘₯ to find 𝑓 of π‘₯. Let’s do this in steps. First, 𝑓 prime of π‘₯ will be an antiderivative of 𝑓 double prime of π‘₯ since when we differentiate 𝑓 prime of π‘₯ with respect to π‘₯, we get 𝑓 double prime of π‘₯. Therefore, we can find an expression for 𝑓 prime of π‘₯ by integrating 𝑓 double prime of π‘₯ with respect to π‘₯. This will be up to a constant of integration 𝐢.

We can then substitute the expression we’re given in the question for 𝑓 double prime of π‘₯. We have 𝑓 prime of π‘₯ is the indefinite integral of six π‘₯ with respect to π‘₯ up to a constant of integration. We could then evaluate this integral by using the power rule for integration. We recall this tells us for any real constants π‘Ž and 𝑛, the integral of π‘Žπ‘₯ to the 𝑛th power with respect to π‘₯ is π‘Ž times π‘₯ to the power of 𝑛 plus one divided by 𝑛 plus one plus a constant of integration 𝐢. We add one to the exponent of π‘₯ and then divide by this new exponent. To apply this to six π‘₯, we recall we can rewrite π‘₯ as π‘₯ to the first power. We add one to this exponent of one to get a new exponent of two and then divide by the new exponent of two. This gives us six π‘₯ squared over two. And we add a constant of integration 𝐢.

And simplifying this, we’ve shown 𝑓 prime of π‘₯ is equal to three π‘₯ squared plus the constant of integration 𝐢. But we’re not trying to find 𝑓 prime of π‘₯. We’re trying to find 𝑓 of π‘₯, so we’ll integrate this one more time. Since the derivative of 𝑓 of π‘₯ with respect to π‘₯ is 𝑓 prime of π‘₯, 𝑓 of π‘₯ is an antiderivative of 𝑓 prime of π‘₯, so we can find an expression for 𝑓 of π‘₯ by integrating 𝑓 prime of π‘₯ with respect to π‘₯. That’s the integral of three π‘₯ squared plus 𝐢 with respect to π‘₯. And once again, this is a polynomial expression, so we can integrate this term by term by using the power rule for integration. In our first term, we add one to our exponent of two to get a new exponent of three and divide by this new exponent. We get three π‘₯ cubed over three, and we’ll add a constant of integration at the end of our expression.

To integrate our second term, we can use the power rule for integration. However, we can also use the derivative of 𝐢π‘₯ with respect to π‘₯ will be equal to 𝐢 because 𝐢 is a constant. In other words, 𝐢 times π‘₯ is an antiderivative of 𝐢. Finally, we need to add a constant of integration to this expression. We’ll call this 𝐷. Simplifying this expression gives us that 𝑓 of π‘₯ is equal to π‘₯ cubed plus 𝐢π‘₯ plus 𝐷 for some constants 𝐢 and 𝐷. But this is not enough to answer our question because we need to find the entire equation of the curve. We need to find the values of 𝐢 and 𝐷. To do this, we note that we have an equation involving the variable 𝐢: 𝑓 prime of π‘₯ is equal to three π‘₯ squared plus 𝐢.

And remember, if we graph the function 𝑦 is equal to 𝑓 of π‘₯, then 𝑓 prime of π‘₯ represents the slope of this graph at π‘₯. And we’re given an equation for the tangent to the graph at a point, and we can use this to find the slope of the curve at a point. We’ll rearrange the equation of the tangent. Doing this gives us 𝑦 is equal to six π‘₯ plus eight, and we can note that this is the slope–intercept form of a line. Therefore, the slope of the tangent line to the curve at the point negative two, negative four is six.

But remember, the slope of the tangent to our curve is the same as the slope of our curve. In other words, 𝑓 prime evaluated at negative two must also be equal to six. And if 𝑓 prime of negative two is equal to six, we can substitute negative two into our equation for 𝑓 prime of π‘₯. This then gives us that six is equal to three times negative two squared plus 𝐢. And we can then solve this equation for 𝐢. We have six is equal to 12 plus 𝐢, and then we subtract 12 from both sides of the equation to get that 𝐢 is equal to negative six.

But we’re not done yet. We also need to find the value of 𝐷. And to do this, we’re going to need to use the fact that there is a tangent line to our curve at the point negative two, negative four. And in particular, this means that the point negative two, negative four lies on our curve 𝑦 is equal to 𝑓 of π‘₯. Therefore, 𝑓 evaluated at negative two is equal to negative four. We can substitute π‘₯ is equal to negative two into our equation for 𝑓 of π‘₯. And we’re also going to use the fact that 𝐢 is negative six. Substituting these values into the equation, we get negative four is equal to negative two cubed minus six times negative two plus 𝐷.

And now we just solve this equation for 𝐷. We get negative four minus eight plus 12 is equal to 𝐷. And then we subtract four from both sides of the equation and simplify. We get 𝐷 is equal to negative eight. Finally, all we need to do is substitute 𝐷 is equal to negative eight into our function 𝑓 of π‘₯. This gives us that 𝑓 of π‘₯ is equal to π‘₯ cubed minus six π‘₯ minus eight. But remember, we’re asked to find the equation of the curve. So we need to write this as 𝑦 is equal to π‘₯ cubed minus six π‘₯ minus eight, which is our final answer.

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