Find the equation whose graph is a parabola with vertex (2,4), vertical axis of symmetry, and contains the point (1,-1). Express your answer in the form "ax^2+bx+c".
Parabola
vertex (2,4), point (1,-1)
Find the equation.
Hello Guest!
\( y-y_v=a(x-x_v)^2\\ y-4=a(x-2)^2\\ y=a(x-2)^2+4\\ -1=a(1-2)^2+4\\ a=\dfrac{-1-4}{(1-2)^2}\\ \color{blue}a=-5\)
\(y=a(x-2)^2+4\\ y=-5(x-2)^2+4\\ y=-5(x^2-4x+4)+4\\ y=-5x^2+20x-20+4\\ y=ax^2+bx+c\\ \color{blue}y=-5x^2+20x-16\)
!