Answer:
[tex](f-g)(x)=x^2-6x+5[/tex] dom: (-∞,∞)
[tex](\frac{f}{g})(x)=x-7+\frac{7}{x+2}[/tex] dom: (–∞,–2)u(–2,∞)
Step-by-step explanation:
[tex](f-g)(x)=f(x)-g(x)=(x^2-5x+7)-(x+2)[/tex] *distribute the negative sign into (x+2)!
[tex]x^2-5x+7-x-2\\x^2-6x+5[/tex]
a parabola (anything that begins with [tex]x^2[/tex]) will have a domain of (-∞,∞) or all real numbers!!
[tex](\frac{f}{g} )(x)= \frac{x^2-5x+7}{x+2}[/tex] use synthetic division to divide (the attached picture)
domain: Because the graph is not continuous, you have to write the domains on both sides of the asymptote which is (-∞,-2)u(-2,∞)
synthetic division:
1. take the divisor (x+2) and solve for x. x= –2 this goes in the top left corner
2. write the numbers AND their signs on the top row. if there is no number and just the variable (like [tex]x^2[/tex] ) just write 1.
3. the first number gets pulled down
4. multiply -2 by 1 and subtract it from –5. (-5-2= -7)
multiply -2 by -7 and add that to the next number in the top row which is -7. (-7 + 14=7)
5. the first number in the bottom row of numbers is the first number in the answer but with one less exponent than the dividend. **write 1 as x**
6. the last number in the bottom row, if it is not 0, is a remainder. write it as that number over the divisor. in this case the remainder is 7. so write it as [tex]\frac{7}{x+2}[/tex]
