What are #a# and #k# so the points #(1,15)# and #( 2,- 5)# are on the graph #y = a ( x + 1) ^ { 2} + k#?

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Answer 1 · 2017-11-03T18:56:32

#a =-4# and #k = 31#.

We write a system of equation

#15 = a(1 + 1)^2 + k#

#15 = 4a + k#

Now

#-5 = a(2 + 1)^2 + k#

#-5 = a(9) + k#

#-5 = 9a + k#

So if we substitute the second equation into the first

#15 = 4a + (-5 - 9a)#

#15 = 4a - 5 - 9a#

#20 = -5a#

#a = -4#

Now solve for #k#.

#15 = 4(-4) + k#

#15 + 16 = k#

#k = 31#

Hopefully this helps!