Find the quadratic function yequalsf(x) whose graph has a vertex (negative 3,3) and passes through the point (negative 6,0). Write the function in standard form.
\(\text{From the vertex we have}\\ (y-3) = a(x+3)^2\\ \text{Now from the point}\\ (0-3) = a(-6+3)^2\\ -3 = 9a\\ a=-\dfrac 1 3\\ y = -\dfrac 1 3 (x+3)^2 + 3\)
Thanks, Rom.....
In standard form we have
y = (-1/3) (x+ 3)^2 + 3
y = (-1/3) (x^2 + 6x +9) + 3
y = (-1/3)x^2 - 2x - 3 + 3
y = (1/3)x^2 - 2x
