# How to find coefficients of parabola based on graph?

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: • you have $$f(2)=5$$ and $$f(-2)=1$$ – Dr. Sonnhard Graubner Jul 1 '17 at 13:32
• Why is $c=-3$. ? – Archis Welankar Jul 1 '17 at 13:33
• @Dr.SonnhardGraubner $f'$ – Shuri2060 Jul 1 '17 at 13:33
• It seems I missed that, tnx now that s easy – eugene_sunic Jul 1 '17 at 13:34
• the plot says $g$. is $g=f'$ or $g=f$ or $g=f''$ or $g= ???$ – Dando18 Jul 1 '17 at 13:34

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 $f(x) = \frac 1 2 x^3 + \frac 1 2 x^2 - 3x + c$

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

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.

• Simple and perfect – eugene_sunic Jul 1 '17 at 13:43