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I have a quadratic graph.

I know that I have the following options:

Vietes formula, vertex (x0,yo), the quadratic equation to get x1,x2. Somehow I don't see how I would get the coefficients A and B. The C coefficient is C=3 as it can be read from the graph easily.

How should I tackle the problem?

Just to note that this graph is the function derivative of a previous graph, in other words

$${f(x)'}=Ax^2+Bx-3$$

The problems says that I have to find $$f(x)$$ I know that I have to integrate and therefore easily get the equation but I need the coefficients.

Image of the derivative graph: enter image description here

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    $\begingroup$ you have $$f(2)=5$$ and $$f(-2)=1$$ $\endgroup$ – Dr. Sonnhard Graubner Jul 1 '17 at 13:32
  • $\begingroup$ Why is $c=-3$. ? $\endgroup$ – Archis Welankar Jul 1 '17 at 13:33
  • $\begingroup$ @Dr.SonnhardGraubner $f'$ $\endgroup$ – Shuri2060 Jul 1 '17 at 13:33
  • $\begingroup$ It seems I missed that, tnx now that s easy $\endgroup$ – eugene_sunic Jul 1 '17 at 13:34
  • $\begingroup$ the plot says $g$. is $g=f'$ or $g=f$ or $g=f''$ or $g= ???$ $\endgroup$ – Dando18 Jul 1 '17 at 13:34
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You have the equation with two unknowns $f'(x)=Ax^2 + Bx - 3$ and from the graph you know $(2,5)$ and $(-2,1)$ are points on $f$. So you have a system of equations,

\begin{align*} 5 &= A(2)^2 + B(2) - 3 \\ 1 &= A(-2)^2 + B(-2) - 3 \end{align*}

So $A=\frac 3 2$, $B=1$ meaning $f'(x)=\frac 3 2 x^2 + x - 3$ and $f(x) = \frac 1 2 x^3 + \frac 1 2 x^2 - 3x + c$

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you will get $$4a+2b=8$$ and $$4a-2b=4$$ form here we get $$a=\frac{3}{2}$$ and $$b=1$$

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You have $f'(-2)=1$, $f'(0) = -3$, $f'(2)= 5$.

So $$f'(-2)=1=A(-2)^2+2B(-2)-3$$ $$f'(2)= 5=A(2)^2+B(2)-3$$

and you can solve.

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  • $\begingroup$ Simple and perfect $\endgroup$ – eugene_sunic Jul 1 '17 at 13:43

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