Answer:
First let see if we can simplify the piecewise.
[tex] \frac{x {}^{2} - 4 }{x - 2} = \frac{(x + 2)(x - 2)}{x - 2} = (x + 2)[/tex]
So rewrite the top equation as x+2.
Second, let evaluate the equation.
For a function to be continuous, one sided limits must be equal to each other so
Limit as of x+2 approaches two, must equal the limit as x approaches ax^2-bx-14
We direct subsitue for both functions.
[tex](2 + 2) = 4[/tex]
[tex]a( {2)}^{2} - b(2) - 14[/tex]
[tex]4a - 2b - 14 = 4[/tex]
[tex]4a - 2b - 18[/tex]
Also the limit as x approaches 3 of 10x-a+b must equal the limit of ax^2-bx-14 as x approaches 3.
So we direct subsitue again
[tex]a(3) {}^{2} - b(3) - 14[/tex]
[tex]9a - 3b - 14[/tex]
[tex]10(3) - a + b[/tex]
[tex]30 - a + b[/tex]
Set them equal to each other
[tex]9a - 3b - 14 = 30 - a + b[/tex]
[tex]10a - 4b - 44[/tex]
Now,we have a system of equations
[tex]10a - 4b = 44[/tex]
and
[tex]4a - 2b = 18[/tex]
Eliminate b variable.
[tex] - 2(4a - 2b = 18) = - 8a + 4b = - 36[/tex]
Add both equations so we get
[tex]10a - 4b = 44[/tex]
[tex](10a - 4b = 44 )+ ( - 8a + 4b = - 36)[/tex]
[tex]2a = 8[/tex]
[tex]a = 4[/tex]
Plug back in to solve for b.
[tex]10(4) - 4b = 44[/tex]
[tex]40 - 4b = 44[/tex]
[tex] - 4b = 4[/tex]
[tex]b = - 1[/tex]
So a=4 ,b=-1
