Science:Math Exam Resources/Courses/MATH110/December 2011/Question 07 (d)

MATH110 December 2011

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Question 07 (d)
Let $f(x)=x^{2}-6x$ and let P be the point (4, -12). d) Find the slopes of all lines through P which are also tangent to the curve $y=f(x)$.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Suppose we have a tangent line at a point on the curve, and a line between the same point on the curve and P. If the two lines are equal, what is true about their slopes?

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. This question is asking which lines through $P$ are also tangent lines to the curve. In order for a line through P to be a tangent line, it must pass through a point on the curve. We found the slope of such a line in part (b). Furthermore, the slope between that point and P must be equal to the value of the derivative at that point. which is what we found in part (c). We know that the slope of a line between P (4, -12) and a point (a, a2 - 6a) on the curve is: $-{\frac {a^{2}-6a+12}{4-a}}$ The slope of the tangent line at the same point on the curve, (a, a2 - 6a) is: $\displaystyle 2a-6$ We set these two slopes equal to each other and solve for a. ${\begin{aligned}-{\frac {a^{2}-6a+12}{4-a}}&=2a-6\\-(a^{2}-6a+12)&=(2a-6)(4-a)\\-a^{2}+6a-12)&=-2a^{2}+14a-24\\a^{2}-8a+12)&=0\\(a-6)(a-2)&=0\end{aligned}}$ So a = 2 and a = 6. Plugging these back into our original slopes (either one will work, since we have found the values where the two are equal) we get the two slopes of 6 and -2.

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