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Given a 3. degree function:

$f(x) = ax^3 + bx^2 + cx + d$

Furthermore, I am given

  • the graph of function $f(x)$ passes through x-axis at origin.
  • the tangent at point P(-3, 0) is parallel with line $y = 6x$

How can I solve for a, b, c and d ?

So far I have,

Since graph passes through origin, $f(0) = 0$ so, $d = 0$ ----------- (1)

Graph also passes through P(-3, 0) so I get,

$0 = -27a + 9b -3c + d$ ----------- (2)

Now to get tangent of line at P(-3, 0) taking first derivative w.r.t $x$ I get,

$f'(x) = 3ax^2 + 2bx +c$

Since tangent is parallel to $y = 6x$, we get slope $m = 6$

$6 = 27a - 6b + c$ ----------- (3)

How can I get fourth equation to solve for all the values.

EDIT 2: Condition 1 (first bullet point), in German:

 - Die Graph von f berührt die x-Achse im Ursprung

EDIT: Could the catch be in the first condition where it explicitly mentions that curve passes the 'x-axis' at origin. Giving another condition for curvature/maxima of the curve. Just my assumption here.

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  • $\begingroup$ Tangent also passes through (-3,0). $\endgroup$ – StubbornAtom Mar 28 '17 at 19:23
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    $\begingroup$ You are doing right. You can't find $a,b,c$, but you can put $b$ and $c$ as a function of $a$ and find the general result. $\endgroup$ – Arnaldo Mar 28 '17 at 19:24
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    $\begingroup$ I don't think you have enough information. Are you sure you have stated the problem correctly? As is, the answer is a one parameter family of cubics. $\endgroup$ – Ethan Bolker Mar 28 '17 at 19:24
  • $\begingroup$ @StubbornAtom The OP has already used that fact since the curve goes through $(-3,0)$ at that point of tangency. $\endgroup$ – Ethan Bolker Mar 28 '17 at 19:26
  • $\begingroup$ @EthanBolker well the original question was in German so I might have lost some information there. However, My guess so far is that the information given in first condition that graph passes the 'x-axis' at origin gives two different condition. 1. it passes through origin. 2. it gives concavitiy of the graph. could it be? If so I can get another equation and solve it. $\endgroup$ – ro ko Mar 28 '17 at 19:30
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So credit goes to this answer from mathlounge.de and all the comments with discussion.

Apparently for the fourth equation, Since f passes/touches x-axis at origin, f has a horizontal tangent line with slope = 0. This gives, $f'(0) = 0$

which means

$c = 0$

and the rest can be solved now. Thanks guys for all the comments.

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