x^2... sorry
y = 10x^2 + bx + c
first, work with y'
y' = 20x + b
as the vertex is (-8;-9) which we southerners call the turning point, then y' = 0 where x = -8
so 0 = 20(-8) + b
0 = -160 + b
b = 160
now just work out c
y = 10x^2 + 160x + c
sub (-8;-9)
-9 = 10(64) + 160(-8) +c
-9 = 640 - 1280 + c
631 = c
y = 10x^2 + 160x + 631
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Verified answer
y = 10x^2 + bx + c
first, work with y'
y' = 20x + b
as the vertex is (-8;-9) which we southerners call the turning point, then y' = 0 where x = -8
so 0 = 20(-8) + b
0 = -160 + b
b = 160
now just work out c
y = 10x^2 + 160x + c
sub (-8;-9)
-9 = 10(64) + 160(-8) +c
-9 = 640 - 1280 + c
631 = c
y = 10x^2 + 160x + 631