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)
