Question

select the correct answer from each drop-down menu.

f(x) = x² - 8x + 15

g(x) = x - 3

h(x) = f(x) + g(x)

h(x) = dropdown . the domain of h(x) is dropdown ∪ dropdown .

reset next

Explanation:

Step1: Find h(x)

Substitute \( f(x) = x^2 - 8x + 15 \) and \( g(x) = x - 3 \) into \( h(x)=f(x)+g(x) \).
\( h(x)=(x^2 - 8x + 15)+(x - 3) \)
Simplify the expression: \( x^2 - 8x + x+15 - 3=x^2 - 7x + 12 \)

Step2: Analyze the domain

First, note that if we consider the original functions, but when we add them, \( f(x) \) is a polynomial (domain all real numbers) and \( g(x) \) is a polynomial (domain all real numbers). Wait, but maybe there was a mis - type and originally it was a rational function? Wait, no, the given \( f(x)=x^2 - 8x + 15 \), \( g(x)=x - 3 \). Wait, maybe the original problem was a quotient? But as per the given \( h(x)=f(x)+g(x) \), both \( f(x) \) and \( g(x) \) are polynomials. The sum of two polynomials is a polynomial, and the domain of a polynomial is all real numbers \( (-\infty,\infty) \). But maybe there was a mistake in the problem statement, and it was supposed to be \( h(x)=\frac{f(x)}{g(x)} \)? Let's check: If \( h(x)=\frac{f(x)}{g(x)}=\frac{x^2 - 8x + 15}{x - 3} \), we can factor \( f(x) \): \( x^2 - 8x + 15=(x - 3)(x - 5) \), so \( h(x)=\frac{(x - 3)(x - 5)}{x - 3}=x - 5 \), with the restriction \( x
eq3 \) (since we can't divide by zero).

Assuming that there was a typo and \( h(x)=\frac{f(x)}{g(x)} \) (because the domain has a union, which suggests a restricted domain):

Step1 (corrected, assuming quotient):

Factor \( f(x) \): \( x^2-8x + 15=(x - 3)(x - 5) \)
So \( h(x)=\frac{(x - 3)(x - 5)}{x - 3} \), cancel out \( x - 3 \) (for \( x
eq3 \)) to get \( h(x)=x - 5 \) (with \( x
eq3 \))

Step2 (corrected, assuming quotient):

The domain of \( h(x)=\frac{f(x)}{g(x)} \) is all real numbers except \( x = 3 \). So the domain is \( (-\infty,3)\cup(3,\infty) \)

Answer:

If \( h(x)=f(x)+g(x) \), \( h(x)=x^2 - 7x + 12 \), domain \( (-\infty,\infty) \)
If \( h(x)=\frac{f(x)}{g(x)} \) (more likely due to domain with union), \( h(x)=x - 5 \) (for \( x
eq3 \)), domain \( (-\infty,3)\cup(3,\infty) \)

Since the domain has a union, we assume the problem was a quotient. So:
\( h(x)=x - 5 \) (with \( x
eq3 \)), domain \( (-\infty,3)\cup(3,\infty) \)