Find the equation whose graph is a parabola with vertex (2,4), vertical axis of symmetry, and contains the point (1,-11). Express your answer in the form "ax^2+bx+c".
The general equation of a standard parabola is \(f(x)=ax^2+bx+c\). Given the vertex and another arbitrary point that lies on the parabola, it should be possible to generate the specific equation of this parabola.
If the problem had not given the coordinates of the vertex, then we know that the coordinates of any vertex is given by \(V=\left(-\frac{b}{2a},f\left(-\frac{b}{2a}\right)\right)\). The x-coordinate of the vertex is given as 2.
\(-\frac{b}{2a}=2\\ b=-4a\)
Let's apply this substitution for our arbitrary function.
\(f(x)=ax^2-4ax+c\)
We know two points that lie on this parabola, (2, 4) and (1, -11). Let's use both of these to solve a system of linear equations.
\(f(2)=4a-8a+c\\ \text{Eq1: }-4a+c=4\)
\(f(1)=a-4a+c\\ \text{Eq2: }-3a+c=-11\)
I will use the elimination method to solve this system.
\(\quad -4a+c= \;\,4\\ +(\;\,3a-c=11)\\ \text{______________}\\ \quad -a \quad\quad = 15\\ a=-15\\ b=60\\ c=-56\)
Equation of Parabola: \(f(x)=-15x^2+60x-56\)
