Given #f(x)=x+2, g(x)=x-3, h(x)=x+4# how do you determine #y=(f(x))/(h(x))times(g(x))/(h(x))#?

Question

James Allen · Accepted Answer

#y = frac(x^(2) - x - 6)(x^(2) + 8 x + 16)#

We have: #y = frac(f(x))(h(x)) times frac(g(x))(h(x))#

#Rightarrow y = frac(f(x) times g(x))((h(x))^(2))#

Let's substitute the expressions for #f(x)#, #g(x)# and #h(x)# into the equation:

#Rightarrow y = frac((x + 2) times (x - 3))((x + 4)^(2))#

Then, let's expand the parentheses:

#Rightarrow y = frac(x^(2) + 2 x - 3 x - 6)(x^(2) + 4 x + 4 x + 16)#

#therefore y = frac(x^(2) - x - 6)(x^(2) + 8 x + 16)#

Serenity Jones · Answer

undefined To determine $y = \frac{{f(x)}}{{h(x)}} \times \frac{{g(x)}}{{h(x)}}$, we first find each individual fraction and then multiply them.

Given $f(x) = x + 2$, $g(x) = x - 3$, and $h(x) = x + 4$, the fractions are:

$$\frac{{f(x)}}{{h(x)}} = \frac{{x + 2}}{{x + 4}}$$

$$\frac{{g(x)}}{{h(x)}} = \frac{{x - 3}}{{x + 4}}$$

Multiplying these fractions together:

$$y = \left(\frac{{x + 2}}{{x + 4}}\right) \times \left(\frac{{x - 3}}{{x + 4}}\right)$$

$$= \frac{{(x + 2)(x - 3)}}{{(x + 4)(x + 4)}}$$

$$= \frac{{x^2 - x - 6}}{{x^2 + 8x + 16}}$$

So, $y = \frac{{x^2 - x - 6}}{{x^2 + 8x + 16}}$.