Question

The vertex of this parabola is at (4, -3). When the x-value is 5, the

Answer 1

$The\ vertex\ form:y=a(x-k)^2+h\\\\where\ (k;\ h)\ \ are\ coordinates\ of\ vertex\\------------------------\\y=a(x-4)^2-3\\\\the\ point\ on\ parabola\ (5;-6)\\Put\ x=5\ and\ y=-6\ to\ equation:\\\\a(5-4)^2-3=-6\\a(1)^2=-6+3\\a=-3\\\\Answer:y=-3(x-4)^2-3=-3(x^2-8x+16)-3\\\\=-3x^2+24x-48-3=-3x^2+24x-51$