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So I thought I was doing it right but I can't be. I keep getting a positive slope for the TL and the coords I get don't make sense for the graph of the function.

I took a picture of my work. If someone could please help, I'd greatly appreciate it. Thanks!

enter image description here

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  • $\begingroup$ Thanks for the edit @tootone $\endgroup$ – Cozen Feb 19 '14 at 16:02
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Let the required point on the curve be $(k,k^2-3)$. Then the slope of the tangent at that point equals $2k$. Further we know it passes through two points $(-2,0)$ and $(k,k^2-3)$. So, lets define it by $L:y = 2kx+c$.

Substituting the $2$ points in the equation, we get $c = 4k$ and $k^2+4k+3=0$ which gives $k=-3$ or $-1$. So the required points on the graph are $(-3,6)$ and $(-1,-2)$

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  • $\begingroup$ Thanks guys. Yeah I got (-1,-2), but point P doesn't line up with (-1,-2) so It's confusing. And wouldn't the slope and y-int have to be negative since the line is going downward from left to right? $\endgroup$ – Cozen Feb 19 '14 at 16:09
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Suppose the $x$-coordinate of the point in question is $t$, so the $y$-coordinate is $f(t)=t^2-3$ and the point itself is $(t,t^2-3)$. You can use the fact that the slope of the tangent line there is $f'(t) = 2t$.

Since the line passes through $(t,t^2-3)$ and $(-2,0)$, compute the slope and set it equal to $f'(t)$, then find the $t$ that satisfies the equation:

$$\frac{(t^2 - 3) - 0}{(t) - (-2)} = 2t \Leftrightarrow \frac{t^2-3}{t+2}=2t \Leftrightarrow t^2-3 = 2t^2 + 4t$$

as long as $t\neq -2$. This can be rearranged as $t^2+4t+3=0$, which has solutions $t=-1$ and $t=-3$ (so the caveat $t\neq -2$ can be ignored). Geometrically, you must have $t>-2$, so the point in question is $(t,t^2-3) = \boxed{(-1,-2)}$.

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  • $\begingroup$ How are you dismissing $t=-3$? Geometry doesn't require that $\endgroup$ – Sandeep Thilakan Feb 19 '14 at 16:13
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it looks like you've already solved it! You've worked out x=-1 and put this back into the equation for y=-2x-4 to get the point P=(-1,-2)

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  • $\begingroup$ But how do I show that y=-2x-5? f'(x)=2x and if you plug in (-2,0) into y=mx+b, you get 0=2(-2)+b.. so the y-int = 4? So the slope and y-int from the equation are positive. But from the graph.. it should be negative? $\endgroup$ – Cozen Feb 19 '14 at 16:14
  • $\begingroup$ your equation for y=-2x-4 was found correctly and a tangent of this form has a negative slope if the co-efficient for x is negative. The co-efficient in this case is -2 and so the equation indicates the tangent has a negative slope. As you say, the equation is in the form y=mx+b. The y-intercept is therefore b, which you have found to be -4 which fits with the graph shown. You have also found that m=-2. When m is negative, the slope is negative $\endgroup$ – Lucy Feb 19 '14 at 16:17

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