If f(x) is continuios for all x, find a & b such that....?

Question

f(x) =
ax+3, x>5
8, x=5
x^2+bx+a, x<5

Hi, I would know how to do this question EASILY....but unfortunately I really don't know how to find a and b....so I really cannot graph this....I know you would need to draw this but how would you find a and b...any help please ?

Answer 1

Because they have defined f(x) as a continuous function, you know that there cannot be any breaks in the function. So, at x=5, both the x>5 and x<5 functions also need to have 8 as the solution.

f(x) = ax + 3
substituting in x=5: f(5) = a(5) + 3 = 8
thus, a = 1

f(x) = x^2+bx+a, x<5
substituting in x=5: f(5) = 5^2+b(5)+1 = 8
b = -18/5

Hope that helps...

Answer 2

If there is continuity, the equations in the different sectors must "mesh" at the sector's junctions.
So at x=5,from the right side, ax+3=8 or a=1.
Also at x=5, from the left side, 5^2+5b+1=8. This leads to b=-18/5
Nothing to it really.