the curve with equation y=ax^2+bx+c passes through (0,-1) and has a stationary point of (2,-9). find the values of a, b and c

[tex]F(x)=ax^{2} +bx+c[/tex]
when we derivate the original general equation
we get
[tex]F'(x)=2ax+x[/tex]
For a stationary point we know that the gradient m which in this point is expressed by F'(x) is equal to 0 in the given x coordinate of the stationary point mathematically ⇒F'(2)=0
When we plug in the x coordinate in our derived equation we have
[tex]2a(2)+2=0\\4a+2=0\\4a=2\\\frac{4a}{4} =\frac{2}{4} \\a=\frac{1}{2}[/tex]
Therefore we can plug in the value of a we have got in the original equation.
[tex]F(x)=\frac{1}{2} x^{2} +bx+c\\[/tex]
We have two unknowns in the equation with two points in the function we can use to get the unknowns b and c
In the F(0)=-1
[tex]-1=\frac{1}{2} (0)^{2} +b(0)+c\\c=0[/tex]

But if we are to check carefully we know that c is the y-intercept wherein x=0 and we already had the point so there was no point of calculating it.

To get the value of b now substitute/plug in the other point (2,-9) in the equation
[tex]F(x)=\frac{1}{2} x^{2} +bx\\c=0 \\F(2)=-9\\-9=\frac{1}{2}(2)^{2} +b(2)\\ -9=2+2b\\-9-2=2b\\-11=2b\\b=\frac{-11}{2}[/tex]