Question

Solve (1,21),(2,6);y=a(x+1)^2 +k

Answer 1

Put the coordinates of the points (1, 21) and (2, 6) to the equation of the function:

$y=a(x+1)^2+k\\\\(1,\ 21)\\\\21=a(1+1)^2+k\\21=a(2)^2+k\\21=4a+k\qquad\text{subtract 4a from both sides}\\21-4a=k\qquad(*)\\\\(2,\ 6)\\\\6=a(2+1)^2+k\\6=a(3)^2+k\\6=9a+k\qquad(**)$

Substitute (*) to (**):

$6=9a+(21-4a)\\6=9a+21-4a\\6=5a+21\qquad\text{subtract 21 from both sides}\\-15=5a\qquad\text{divide both sides by 5}\\-3=a\to\boxed{a=-3}$

Put the value of a to (*):

$k=21-4(-3)\\k=21+12\\\boxed{k=33}$

Answer:

$y=-3(x+1)^2+33$