A, C, and D are correct
B:
3f(-1) = 3[(-1)^2 + 3(-1)]
= 3[1 - 3]
= 3(-2) = -6
E)
x^2 + 3x = 4
x^2 + 3x - 4 = 0
(x+4)(x-1) = 0
x = -4 or x = 1
checking for x = -4
f(-4) = (-4)^2 + 3(-4)
= 16-12 = 4 , checks!
try for x = 1 the same way, you should get 4 as well
Question:
Given f(x)=x^2+3x
x such that f(x)=4 (ie y=4, what is x?)
I am honestly very confused by this question to even attempt to do it.
Though here is what I have on the rest of my sheet as my answers. Please correct me if I'm wrong. Thank you!
Given f(x)=x^2+3x
A. f(2)= 10
B. 3f(-1)=6
C. f(0)+f(2)= 10
D. f(t+1)= t^2+5t+4
E. x such that f(x)=4 (ie y=4 what is x?)
Don't have an answer for this one...
2 answers
Oh my god, thank you so so much! :-)