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So we have that $a=1$, $b=0$, $c=1$, so our point reduces to $(t,t^{2}+1)$. Plugging that into the general equation for a line we have $t^{2}+ 1 = mt + b$ where $m$ is the slope and $b$ is the $y$-intercept. So we need this line to intersect $f$ once. Setting the graph of $f$ and the line equal to each other we have $t^{2}+ 1 - b + mt = t^{2}+1$. Hence $t = \frac{b}{m}$. So plugging into the given point we have $(\frac{b}{m},(\frac{b}{m})^{2}+1)$. Plugging that into the line we have $\frac{b^{2}}{m^{2}}+1=\frac{b}{m}m+b$. So $\frac{b^{2}}{m^{2}} - 2b + 1 = 0$. Thus $b = \frac{2 \pm \sqrt{4 - 4(\frac{1}{m^{2}})}}{\frac{2}{m^{2}}}$. I feel like I am going in circles with this one. Does anyone have a suggestion that would make this easier?

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  • $\begingroup$ Is this the entire question? 'Cause there are an infinite number of solutions. For example, no matter what $a$, $b$, and $c$ are, the line $x=t$ satisfies the stated requirement. $\endgroup$ – rogerl Mar 22 '15 at 15:23
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I would utilize the result that a non-vertical line will intersect the graph of $f(x)=x^2+1$ only once, exactly when the line is tangent to the graph of $f$. You can try to prove this.

It doesn't say so in your question that the point $(t,at^2+bt+c)$ is a point on the graph of $f$, but from what you write in your post, this seems to be the case, and that we can let $a=1$, $b=0$, and $c=1$.

Also, it doesn't say so in your post whether the line is non-vertical or not, but I assume that this is what is meant.

Since the line is tangent to the graph at the point $(t,t^2+1)$, the slope of the line must be equal to the slope of the function at this point. This slope is $2t$, so if the line is given by $mx+b$, then $m=2t$. Then you can plug the point $(t,t^2+1)$ into the equation $y=(2t)x+b$ for the line to obtain $b$.

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    $\begingroup$ Plugging that point in we get $t^{2}+ 1 = 2t^{2} + b$. So $b = -t^{2} + 1$. Thus the line is $y = 2tx - t^{2} + 1$. I dont know if that is right. $\endgroup$ – Jack Mar 22 '15 at 15:33
  • $\begingroup$ I agree with this @Tim $\endgroup$ – Mankind Mar 22 '15 at 15:37
  • $\begingroup$ Thanks. I caved in and emailed the professor I TA for. This is for a precalc class and I couldnt figure it out without using calc either. Oh well..gotta grade these finals!!! $\endgroup$ – Jack Mar 22 '15 at 15:40
  • $\begingroup$ @Tim, sure. It would be interesting to know what the intended solution was. $\endgroup$ – Mankind Mar 22 '15 at 16:30
  • $\begingroup$ Answer posted above $\endgroup$ – Jack Mar 26 '15 at 6:11
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So here is the answer:

The point $(t,t^{2}+1)$ lies on the graph $f(x) = x^{2}+1$ and we need to find the line that passes through this point and intersects with the graph only once. So we can precede as follows:

The line in question is $y = kx+b$. To find the slope we have $ k = \frac{t^{2}+1-y}{t - x}$. We have substitue $y=x^{2}+1$ to attain $k=\frac{t^{2}-x^{2}}{t - x}$, which becomes $x^{2}-kx-t^{2}+kt=0$. We now need to solve for $k$ so that $x$ will only have one solution. This means that the slope we attain will ensure that our line will be tangent to $f$. So we have $x= \frac{k \pm \sqrt{k^{2}- 4(-t^{2}+kt)}}{2}$. We need to set everything in the square root equal to zero for $x$ to have only one solution. Thus $k^{2}-4kt+4t^{2}=0$. So this becomes $(k-2t)^{2}=0$, hence $k=2t$.

So plugging this into our line with the given point we have $t^{2}+1=2t(t) + b$. Solving for $b$ we see that $b=-t^{2}+1$. Thus our line in question which is tangent to $f$ at the given point is $y = 2tx-t^{2}+1$.

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