Answer:
d. c = 100; (x – 10)²
Step-by-step explanation:
Note that (x+a)²=x²+2ax+a².
In our case, 2a=-20, so 'a' must be -10.
Since c=a², and a=-10, c=(-10²)=100.
Since a=-10, the perfect square is (x – 10)²
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1.) f(x)=7(b)^x-2
x=0→f(0)=7(b)^0-2=7(1)-2=7-2→f(0)=5→(x,f(x))=(0,5) Ok
2.) f(x)=-3(b)^x-5
x=0→f(0)=-3(b)^0-5=-3(1)-5=-3-5→f(0)=-8→(x,f(x))=(0,-8) No
3.) f(x)=5(b)^x-1
x=0→f(0)=5(b)^0-1=5(1)-1=5-1→f(0)=4→(x,f(x))=(0,4) No
4.) f(x)=-5(b)^x+10
x=0→f(0)=-5(b)^0+10=-5(1)+10=-5+10→f(0)=5→(x,f(x))=(0,5) Ok
5.) f(x)=2(b)^x+5
x=0→f(0)=2(b)^0+5=2(1)+5=2+5→f(0)=7→(x,f(x))=(0,7) No
Answers:
First option: f(x)=7(b)^x-2
Fourth option: f(x)=-5(b)^x+10
