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This is probaly easy.

Find the equation of the tangent to a curve $y = u (x)$ is $y =(-2a+4)(x-a)+k$, where $k$ is a constant. Given that the curve touches the $x$-axis at $x =2$, find the value of $k$ and the equation of the curve.

peterh
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esoteric elf
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    Hint: That the curve touches the $x$-axis at $x=2$ means that the point $\cal P(2,0)$ is in the curve. So apply the point slope-formula and you will get your result. – Hakim Jun 11 '14 at 15:19
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    $T = {(x,y):y=(−2a+4)(x−a)+k},(2,0) \in T$. So $0 = (-2a+4)(2-a)+k$ and thus $k=...$ – Surb Jun 11 '14 at 15:20
  • the final answer is k=0 – esoteric elf Jun 11 '14 at 15:25
  • and y=-x^2+4x-4, is it i can put a =2? i tried to integrated but find the result quite conflicting – esoteric elf Jun 11 '14 at 15:26
  • I integrated both side, just wonder if a = 2? I don't know why I stumble on this basic stuff, please advice. – esoteric elf Jun 11 '14 at 16:23
  • I'm somewhat confused by the formulation of the question. Do we have a curve given by $u(x)$, where the tangent line at any point of the curve is given by $y=(-2a+4)(x-a)+k$?

    If $a$ is supposed to be a constant, the tangent line to $u(x)$ has the same slope at every point $(x_0,y_0)$ on the curve, which would mean the curve $u(x)$ is actually a straight line. Since it is also given that the curve touches the $x$-axis at $x=2$, this means the tangent has slope $0$ at $x=2$, hence everywhere and so, $u(x)=0$, which doesn't sound right.

     – HSN Jun 11 '14 at 16:36
  • i see why now, thanks – esoteric elf Jun 11 '14 at 17:01
  • @HSN I know I'm too late but this question makes complete sense? The tangent line does not have the same slope at any point. Just look at my answer. – Arbuja Nov 14 '15 at 21:51
  • @user145472 Be sure to choose an answer. – Arbuja Nov 14 '15 at 22:22

2 Answers2

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Letting $y=u(x)$, its derivative is $u'(x)$, and it represents the slope of the tangent to the function $u(x)$ at the point $(x,u(x))$. Thus the line tangent to the function $u(x)$ at the point $c$, is given by the equation $$y-u(c) = u'(c)\cdot (x-c)\to \\y=x\cdot u'(c) +u(c) -c\cdot u'(c)=\\x\cdot (4-2a)+2a^2-4a+k$$ Now you may note that \begin{cases} 4-2a=u'(c)\\ u(c) -c\cdot u'(c) = 2a^2-4a+k \end{cases} so solving the system we obtain \begin{cases} \int 4-2a\, \mathrm{d}c=u(c)\\ u(c) -c\cdot (4-2a) = 2a^2-4a+k \end{cases} \begin{cases} c\cdot (4-2a)+b=u(c), b \text{ is a constant}\\ u(c) -c\cdot (4-2a) = 2a^2-4a+k \end{cases} \begin{cases} c\cdot (4-2a)+b=u(c), b \text{ is a constant}\\ c\cdot (4-2a)+b-c\cdot (4-2a) = 2a^2-4a+k \Rightarrow b=2a^2-4a+k \end{cases} Finally we get to the conclusion that $u(c) = c\cdot (4-2a)+2a^2-4a+k$. Now,$$u(2)= 2\cdot (4-2a)+2a^2-4a+k =0\Rightarrow\\8-4a+2a^2-4a=-k\\k=-2(a^2-4a+4)=-2(a-2)^2\Rightarrow \\u(x) = x\cdot (4-2a) +2a^2-4a-2a^2+8a-8\\\color{green}{u(x) = 2x\cdot (2-a)+4a-8}$$

sirfoga
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I think that @Foga's answer too long.

@Harikim gives you a shortcut but here is the general approach.

First thing to note is that when you take $(2a-4)(x-a)+k$ note that this is in the tangent line form of $f'(a)(x-a)+f(a)$.

So $f'(x)=2x+4$

Then taking the integral $$\int{f'(x)}=\int{2x-4}$$ $$f(x)=x^2-4x+c$$

Now since the curve is supposed to touch at $(2,0)$ we should first check the derivative and make sure is at zero. to make sure there are no tricks to this.

$f'(2)=0$ This we can substitute (2,0) into $y=x^2-4x+c$ $$0=4-8+c$$ $$c=4$$

Now if we take the tangent line formula of $y=f'(a)(x-a)+f(a)$ where $a=2$ to $y=(2a-4)(x-a)+k$. You realize $k=f(a)=f(2)$

So $k=f(2)=0$

Arbuja
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