You have to understand how piecewise functions first I hope you do because I won't be explaining it here
1. if it is continiuous at x=1 then the part of the function that approaches 1 from the left is the same value so we have f(x)=3-x for x<1, so evaluate it for x=1, we would get 3-1=2 so if a=2 and b=3
f(x)=2x²+3x for x=1, we get 2+3=5
it approaches 2 from the left but equals 5 2≠5 so it is not continuous because it doesn't approach the same value at x=1
2. alright so we established that it should approach 2 because we had the f(x)=3-x so f(1)=2=ax²+bx 2=a+b the relationship is 2=a+b
3. same as before, but use the other function, the one that is defined at x≥2 5x-10, 5(2)-10, 10-10, 0 it approaches 0 so we must find one such that f(x)=0 for x=2 with the ax²+bx 0=a(2)²+b(2) 0=4a+2b if we minus 2b both sides -2b=4a divide by 2 -b=2a add b both sides 0=2a+b
4. 2=a+b 0=2a+b
2=a+b minus b both sides 2-b=a subsitute 0=2a+b 0=2(2-b)+b 0=4-2b+b 0=4-b b=4
sub back 2-b=a 2-4=a -2=a
a=-2 b=4
5. graph f(x)=-2x²+4x from x=1 to x=2 and put a closed dot at (1,2) and an open dot at (2,0)
