Given the vertex T(0,1) which closes the x-axes (apsis) with a surface of 5, find distance between x-values on the x-axis?
What I've done: from the formula of the vertex, for x-coordinate; $$\frac{-b^2}{2a}=0 \Rightarrow b=0$$ from the formula of the vertex, for x-coordinate: $$\frac{-b^2-4ac}{4a}=0 \Rightarrow c=-1$$
Now, I have a problem, when I put x=0, for the general equation, The y coordinate does not match the result? $$f(x)=Ax^2+Bx+C \Rightarrow f(0)=A0+0*0-1 \Rightarrow 1 \neq-1$$ I 'm really confused with this, It seems that I've miscalculated c, but how? I Can't figure that out I should be getting 1=1.Why am I not getting it?
Otherwise, I know hot to solve the problems (using Viete formulas) and then plugging x values given by the Viete formulas into my definite Integral with a Surface of 5. I solve for A and get x1 and x2. I hope this is correct?
While waiting a comprehensible translation, I would like to point out that the vertex of the parabola $y=ax^2+bx+c$ is at the point $$\left(-\frac{b}{2a},-\frac{b^2-4ac}{4a}\right).$$ Hence if the vertex is $(0,1)$ then $b=0$ and $c=1$.
I wonder if the 5 means the difference between the $y$-coordinates of the focus and the vertex which is $\frac{1}{4a}$, then we should have $a=1/20$.