Notice that we need to get to a degree fo 6 on the right side.
Since h(x) can't be a degree of 6, (x^2+3x-3)*f(x) has to have a degree of 6.
(x^2+3x-3) has a degree of 2, 6 - 2 = 4
=^._.^=
Hello Pi :))
I really want pie after seeing your pfp.
We know that f(-2) = 3.
So we should make the 2x + 1 = -2 on the graph y = f(2x + 1) + 3.
2x + 1 = -2
2x = -3
x = -1.5
y = f(2x + 1) + 3
y = f(2(-1.5) + 1) + 3
y = f(-2) + 3
f(-2) = 3
y = 3 + 3
y = 6
(-1.5, 6)
I hope this helped.
f(-2) = (-2)^2 - 3
f((-2)^2 - 3) = ((-2)^2 - 3)^2 - 3
t((-2)^2 - 3) = 9 + 2f((-2)^2 - 3)
Nice, but I think you forgot 0, 0^2 = 0.
Happy birthday Cal. :)))
I hope you have a great day today and eats lots and lots of cake.
x - y = 12
y = x - 12
y^2 = 5(x - 2)
(x-12)^2 = 5(x - 2)
Very clear solution. :))
Welcome to web2.0calc.
You're welcome. :))
Plug in x as 2, and solve for k. Once you get k, factor the equation to find the other 2 roots. :))
2x + 7 does not equal 3x + 1, however for some "x" it does.
2x + 7=3x +1
2x + 6 = 3x
6 = x