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I was practicing some basic math and I faced a question that I have no idea what does it requires me to do. Consider this simple function:

$$f(x) = x^2-5x-12$$

Part a and b where pretty easy. I tried to solve part C as you can see below. Unfortunately I cant imagine what has been asked in part D. What does coincident points of intersection mean? (Literally intersection means cutting each other and Coincident means on top of another AFAIK) Any help/explanation would be appreciated!

Part C:

The point at which the line through $(1,-15)$ with slope $m$ cuts the graph of $f(n)$.

$$m=f'(n)=f'(1)=-1$$ $$y-y_1 = m(n-n_1)=>y+15=-1(n-1)$$ $$=>y=-n+1-15$$

Part D:

The values of m such that the points of intersections found in $c$ are coincident.

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Coincident means that the two results you should get in part (c) are the same. Something must be wrong because you only seem to be getting a single result (in fact I don't understand what you're doing there at all). –  Henning Makholm Mar 22 '12 at 18:27
    
Well I am not sure about the part(b) as well...I just tried to find the equation for the tangent line hoping it means the same thing. Any ideas? –  Sean87 Mar 22 '12 at 18:33
    
As far as I can tell, in part c), the slope is given to you. You need to find the two (or possibly one) points where the line intersects the parabola. In d), you need to find the value of $m$ such that there is only one intersection point (this would give the tangent line of course). –  David Mitra Mar 22 '12 at 18:33

1 Answer 1

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The slope of the tangent line to $y=2x^2-5x-12$ at $x=1$ is indeed $-1$, and when $x=1$ we have $y=-15$. However, in solving problems like C), it is best not to notice these facts, because you might be asked a similar question in which the given point is not on the curve. Also, you were asked explicitly to find the intersection points with the curve of the general line with slope $m$ that passes through $(1,15)$, not just the line with specific slpe $m=-1$.

The line through $(1,-15)$ with slope $m$ has equation $y-(-15)=m(x-1)$, that is, $y=mx-m-15$. We want to find the point(s) at which this line meets the curve $y=2x^2-5x-12$.

Substitute $mx-m-15$ for $y$ in the equation of the curve. We get a quadratic equation in $x$, which turns out to be $$2x^2-x(5+m)+(3+m)=0.$$ Solve. In principle, we use the Quadratic Formula, or complete the square. However, the problem poser has kindly chosen numbers that make the quadratic factor nicely. One root is (as we already knew) $x=1$, and the other is $x=\frac{m+3}{2}$. The two roots coincide when $m=-1$.

Remark: The problem is presumably intended as an approach to the tangent line that does not use the derivative. And indeed Fermat, among others, found tangent lines to quite a few curves by using precisely the "double root" idea, well before the work of Newton and Leibniz on the calculus.

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Thanks. Does solving the final equation you give will give us answer to part D? or it is for part C? Sorry I got a bit confused! –  Sean87 Mar 22 '12 at 18:46
    
@Sean87: It is for C). But the two roots are $x=1$ and $x=\frac{m+3}{2}$. So the two roots are equal if $1=\frac{m+3}{2}$. Solve for $m$. We get $m=-1$. –  André Nicolas Mar 22 '12 at 18:56

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