# Question Video: Finding the Points on the Curve of a Cubic Function at Which the Tangent to the Curve Is Parallel to a Given Line Mathematics • Higher Education

Find the points on the curve 𝑦 = 3𝑥³ − 5𝑥 + 7 at which the tangents are parallel to the line 4𝑥 + 𝑦 − 2 = 0.

05:50

### Video Transcript

Find the points on the curve 𝑦 equals three 𝑥 cubed minus five 𝑥 plus seven at which the tangents are parallel to the line four 𝑥 plus 𝑦 minus two equals zero.

First, we recall that two lines are parallel if they have the same slope. We can work out the slope of the line four 𝑥 plus 𝑦 minus two equals zero by first rearranging its equation into the slope–intercept form. Remember the slope–intercept form of the equation of a straight line is 𝑦 equals 𝑚𝑥 plus 𝑏, where 𝑚, the coefficient of 𝑥, represents the slope of the line and 𝑏, the constant term, represents the 𝑦-intercept. To convert the equation of this line into slope–intercept form, we need to subtract four 𝑥 from each side and also add two to each side of the equation. Doing so gives 𝑦 equals negative four 𝑥 plus two. We can now compare this with the slope–intercept form of the equation of a straight line, and we see that the slope of this line, the coefficient of 𝑥, is equal to negative four.

We now know that the tangents that we’re looking for to this curve 𝑦 equals three 𝑥 cubed minus five 𝑥 plus seven must also have a slope of negative four if they are to be parallel to the line four 𝑥 plus 𝑦 minus two equals zero. The slope of a curve is given by its first derivative or slope function, which we refer to as d𝑦 by d𝑥, and can be found by differentiating the equation of the curve with respect to 𝑥. Our function 𝑦 is a polynomial in 𝑥. So we recall the general power rule of differentiation, which tells us that the derivative with respect to 𝑥 of 𝑎𝑥 to the 𝑛th power, where 𝑎 and 𝑛 are real constants, is equal to 𝑎 multiplied by 𝑛 multiplied by 𝑥 to the power of 𝑛 minus one. We multiply by the power and then decrease the power by one.

So let’s now apply this rule to find an expression for d𝑦 by d𝑥. The first term in 𝑦 is three 𝑥 cubed. So differentiating, we have three multiplied by three. Then we reduce the power of 𝑥 by one, giving 𝑥 squared. So we have three multiplied by three 𝑥 squared. To differentiate the term negative five 𝑥, we can think of this as negative five 𝑥 to the first power. So applying the power rule of differentiation, we have negative five multiplied by one multiplied by 𝑥 to the zeroth power. But if we recall that 𝑥 to the power of zero is equal to one, then we don’t need to write this part down.

Finally, we’re differentiating positive seven, and we recall that the derivative of a constant with respect to 𝑥 is equal to zero. To see why this is the case, we can think of the constant seven as seven multiplied by 𝑥 to the power of zero. As we’ve already said, 𝑥 to the power of zero is equal to one. If we were to then apply the general power rule, we’d have seven multiplied by zero multiplied by 𝑥 to the power of negative one. But, of course, multiplying anything by zero just gives zero. So we’re left with three multiplied by three 𝑥 squared minus five multiplied by one, which simplifies to nine 𝑥 squared minus five.

We’ve now found an expression for the slope function of this curve. Remember though that we are looking to determine the points at which the tangents are parallel to the line four 𝑥 plus 𝑦 minus two equals zero, which we have already established are the points where the slope is equal to negative four. To find the 𝑥-coordinates of these points, we can take our expression for the slope function and set it equal to negative four, giving nine 𝑥 squared minus five equals negative four. We’ll now solve this equation to determine the values of 𝑥. First, we add five to each side of the equation, giving nine 𝑥 squared is equal to one. Then we divide by nine, giving 𝑥 squared is equal to one-ninth.

To solve for 𝑥, we take the square root, 𝑥 is equal to plus or minus the square root of one-ninth. This is equivalent to plus or minus the square root of one over the square root of nine. And as these are both square numbers, this simplifies to plus or minus one over three or plus or minus a third. We’ve now found the 𝑥-coordinates of the points on this curve where the tangents are parallel to the given line. The final step is to also find their 𝑦-coordinates, which we can do by substituting each of our 𝑥-values into the equation of the curve. When 𝑥 is equal to positive a third, first of all, 𝑦 is equal to three multiplied by a third cubed minus five multiplied by a third plus seven. That’s three over 27 minus five over three plus seven, which simplifies to 49 over nine.

We could show this by canceling the fraction three over 27 to one over nine and then writing each of the other two terms also as fractions with denominators of nine. When 𝑥 is equal to negative one-third, 𝑦 is equal to three multiplied by negative one-third cubed minus five multiplied by negative one-third plus seven. That’s equal to negative three over 27 plus five over three plus seven, which is equal to 77 over nine. We have therefore found the coordinates of two points on the curve 𝑦 equals three 𝑥 cubed minus five 𝑥 plus seven at which the tangents are parallel to the line four 𝑥 plus 𝑦 minus two equals zero. Those points are one-third, 49 over nine and negative one-third, 77 over nine.