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I'm in an American high school but my teacher gave me some IBDP questions and I'm struggling to solve one of them:

Find the exact values of $m$ for which the line $y= mx+3$ is tangent to the curve with equation $y=3x^2 -x +5$

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    $\begingroup$ Have you learned about derivatives yet? $\endgroup$ – Samasambo Aug 29 '16 at 15:34
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    $\begingroup$ Yes, but only myself. I know the derivative would be 6x-1 for the quadratic. How does that help? $\endgroup$ – Daniel F. Aug 29 '16 at 15:36
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What does it mean for $y=mx+3$ to be tangent to $y=3x^2-x+5$? It means the quadratic equation $3x^2-x+5-(mx+3)=0$ has exactly one solution, or equivalently, the discriminant is zero, so...

Edit: as per @Glen O's comment, I find it necessary to clarify a little bit how my method works.

Let $f(x)$ denote the quadratic function. Then $f'(x)$ (the slope) is a nonzero linear function, so is strictly monotone, or more generally, is injective on $\Bbb R$. Now given a line $l$ whose slope is $k$, suppose it intersects the quadratic function curve at two distinct points $a,b$ with $a<b$, then by Lagrange MVT, there exists $c\in (a,b)$ such that $f'(c)=k$. But $f'(x)$ is injective, so neither $f'(a)$ nor $f'(b)$ equals $k$, so $l$ must not be tangent to $y=f(x)$.

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  • $\begingroup$ Do I use the 3x^2 + x(1-m) +2 = 0 and use the discriminant to solve it equal to 0? $\endgroup$ – Daniel F. Aug 29 '16 at 15:35
  • $\begingroup$ @Matthew yes of course. In this way you get a quadratic equation in terms of $m$, then solve it and you get the desired results. $\endgroup$ – Vim Aug 29 '16 at 15:36
  • $\begingroup$ I tried that but then I got a very complicated answer in the end... Hm, I'll try again. $\endgroup$ – Daniel F. Aug 29 '16 at 15:37
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    $\begingroup$ @Matthew it's not that complicated. In fact, $$\Delta = (m+1)^2-24=0.$$ So you get $m+1=\pm 2\sqrt 6.$ PS you made a mistake in your calculation, the coefficent of the linear term should be $-(1+m)$ instead of $1-m$. $\endgroup$ – Vim Aug 29 '16 at 15:40
  • $\begingroup$ Thank you! I did receive something similar in the end. My question would only be, why did your $(m+1)^2$ become positive when you put it into a square? I do know squared numbers become positive but I thought the "-" sign would stay outside the brackets despite squaring it... $\endgroup$ – Daniel F. Aug 29 '16 at 15:44
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We need to find $x$ values for which the slopes are the same, and the $y$ values are the same. That's what tangent means. Let $f(x) = mx + 3$ and let $g(x) = 3x^2 - x + 5$.

First take the derivative of $g(x)$, which is $$ g'(x) = 6x - 1. $$ This will give you the slope at any given point.

We want $m$ to be equal to $g'(x)$, so that they have the same slope. We also want $f(x)$ to equal $g(x)$. So we have the two equations: $$ m = 6x - 1 $$ and $$ mx + 3 = 3x^2 - x + 5. $$ We can substitute $6x - 1$ for $m$ in the second equation $$ (6x - 1)x + 3 = 3x^2 - x + 5 $$ and solve for $x$.

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OK so I started off by setting the equations equal to each other, which gave me:

$mx + 3 = 3x^2 -x +5$

I put the 3 on the other side:

$mx = 3x^2 -x +2$

Then I set it equal to zero:

$3x^2 + x(1-m) +2 = 0$

And I'm stuck now. I tried using the discriminant or the quadratic formula but I got really complicated answers and they didn't seem right. I tried solving it for the past 30 minutes and I came up with no answer.

Any input would be appreciated.

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  • $\begingroup$ You can typeset equations by surrounding them with dollar signs, like $mx+3$ and centered equations like $$mx+3$$. You made an error on the third line by moving $mx$ to the other side but getting the wrong sign. Set the determinant $B^2-4AC=0$ to get your answer, where $B=1-m$, $A=3$ and $C=2$. $\endgroup$ – Frenzy Li Aug 29 '16 at 15:37
  • $\begingroup$ Thank you for the input! Would it therefore be $3x^2 - (1+m) +2$ ? $\endgroup$ – Daniel F. Aug 29 '16 at 15:39
  • $\begingroup$ Second term involves $x$ but you get the right idea. $\endgroup$ – Frenzy Li Aug 29 '16 at 15:41
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    $\begingroup$ Just a typo, Mathew: $3x^2-(1+m)\bf x+2 = 0$ $\endgroup$ – Namaste Aug 29 '16 at 15:41
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    $\begingroup$ It would have been better to have your work in the original posting, not as an incomplete answer. $\endgroup$ – Teepeemm Aug 29 '16 at 20:51

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