1

You have a quadratic function: $ax^2 + bx + c = y$. If you know $b$ and $c$, are able to plug any domain value $x$ into this blackbox equation and receive a range value $y$, and do not know the vertex of the equation, how could you find the unknown variable $a$.

Hans Engler
  • 15,439
Paul T.
  • 13
  • 1
    I can't tell what you're asking. So you're saying you know the values of $b$, $c$, and $f(x)$ for some $x$? And with these, you want to find $a$? –  May 21 '14 at 00:01
  • Do you mean $ax^2 + bx + c = 0$, or is there a $y$ variable? – recursive recursion May 21 '14 at 00:02
  • If I read this correctly, if you know $b$ and $c$ and if you can plug in any value of $x$ and receive the corresponding value of $y$, then it really becomes a problem with one unknown, namely $a$. – dmk May 21 '14 at 00:10
  • 1
    The solution for $a$ is $\dfrac{y-bx-c}{x^2}$, if that's what you want. –  May 21 '14 at 00:13
  • @SanathDevalapurkar : You should post as an answer. – Patrick Da Silva May 21 '14 at 00:31
  • Hello and welcome to math.stackexchange! I have edited your question a bit - please check if these edits are OK. As to you question" If you can plug in $x$ on the left hand side and are told $y$ on the right hans side, you can set up a single equation for the unknown $a$ and solve for it. Try it with $b = 2, c = 3, x = -2, y = 1$. What is the equation you will get for $a$ and what is the solution? And what is the general process? – Hans Engler May 21 '14 at 00:32
  • @SanathDevalapurkar I will try that thanks. You should post it as an answer. – Paul T. May 21 '14 at 00:34
  • @HansEngler thanks thats great, and thank you for the welcome! – Paul T. May 21 '14 at 00:36
  • @dmk yes I'm looking for a using values b, c, x, and y. – Paul T. May 21 '14 at 00:37
  • @SanathDevalapurkar thank you! So far it seems to work. Please post it as an answer. – Paul T. May 21 '14 at 00:44
  • @PaulT.: What do you mean by "so far it seems to work"? Either you understand why it works and can convince someone else that it does, or you don't and shouldn't rely on 'experience' to confirm correctness. You must verify it yourself. Try $x=0$ in his formula and see if it works! To get the answer yourself, substitute $x=1$ into your blackbox and see what you are supposed to get. – user21820 May 21 '14 at 01:15
  • @user21820 I works except the equation x cannot equal zero for obvious reasons. – Paul T. May 21 '14 at 01:21
  • 1
    @PaulT.: Yes, but do you understand it completely and can you get the answer on your own? The answer itself is unimportant, but knowing why it is the answer is important. – user21820 May 21 '14 at 01:38
  • @user21820 I believe I do a thought process showing how the equation was derived would be helpful. – Paul T. May 21 '14 at 01:57

1 Answers1

2

The solution for $a$ is $$a=\begin{cases}\dfrac{y-bx-c}{x^2} & x\neq0\\ \text{unknown value} & x=0\end{cases}$$ How that is derived can be seen by rearranging and dividing.