<p>f(x)= ax^2+x+c
the line y=3x-1 is tangent to the graph of f(x) at the point 2,1. d^2f/dx^2(1)=0.</p>
<p>find a b and c</p>
<p>dunnno what to do?</p>
<p>calc buffs where are u?</p>
<p>Okay, assuming that:</p>
<p>f(x) = ax^2 + bx + c </p>
<p>here's what I think. Correct me if I'm wrong, this seems like a weird problem to me.</p>
<p>First of all, I think this is a bit of a trick, because the second derivative cannot equal zero with an x^2 term:</p>
<p>f(x) = x^2
f'(x) = 2x
f"(x) = 2</p>
<p>or, in this case:</p>
<p>f(x) = a<em>x^2
f'(x) = a</em>2x
f"(x) = a*2</p>
<p>therefore, if d^2y/dx^2 = f"(x) = 0 = 2a, then a = 0</p>
<p>So now you're down to:</p>
<p>f(x) = bx + c</p>
<p>bx + c has to be a constant slope, which means that any tangent line would essentially be the line you're talking about, so:</p>
<p>f(x) = 3x-1</p>
<p>a = 0
b = 3
c = -1</p>
<p>(Sorry if I've missed something obvious but that's all I could come up with.)</p>
<p>Edit: Okay, this problem doesn't work for me. Something is wrong... Are you sure that's the given info? Because the point (2,1) doesn't exist on the line y = 3x - 1 , as far as I can tell...</p>
<p>ok.</p>
<p>ax^2+x+c -> 2ax+1 -> 2a = 0
a = 0</p>
<p>now we get:</p>
<p>x+c = f(x)
f'(x) = 1.... = 3</p>
<p>at this point, I have to ask you, Where the He11 is b?
is it f(x) = ax^2 + bx + c?</p>
<p>then you would have</p>
<p>3 = 1 + b
b = 2</p>
<p>now we have f(x) = 2(2) + c = 1
f(x) = 2x - 3</p>
<p>but that is prolly wrong</p>
<p>i think a set of values of a, b, and c does not exist that fits these conditions. since f" is 0 at x=1, and f is a quadratic, a=0. this means that the f is a line. and so the only line fitting the condition of the tangent line, i guess, would be the tangent line. therefore f(x)=3x-1. but 3x-1 doesn't go through the point (2,1).</p>
<p>I think you typed this problem in wrong. Look at it again.</p>
<p>yea my bad it was supposed to be ax^2+bx+c</p>
<p>No, there's more wrong than that =P</p>
<p>I assumed the bx part and still doesn't work...</p>
<p>and its supposed to be at 1,2 wow i really butchered this problem</p>
<p>Then I think f(x) does indeed equal your tangent line =)</p>
<p>yea tricky bastards</p>