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# Week 15 in PreCalc 11

One thing I’ve learned this week is about multiplying rational expressions.

It doesn’t really need a lot of explanation as it’s just your simple multiplying of fractions, however with variables this time.

Let’s have a recap:

$(\frac{2}{5}) (\frac{10}{8}) = \frac{20}{40}$

We just multiply straight across, but remember we need to simplify our answer.

20/40 = 1/2

But we can also do it this way:

It’s like prime factorization method. So, you take out all of the prime factors, then just cancel out. Well, that’s exactly what we’re going to do! But let’s add variables to the fun.

So we have this not-so-pretty looking fractions and we might be thinking, how do we solve that?!

Well, I got good news! They’re pretty easy to solve as long as you’re good at factoring!

So first, factor everything.

Now that you’ve factored everything, just cancel out same factors, just like what we did earlier in that simple fraction.

Next – don’t forget about this! Remember the rule that you cannot have a zero on your denominator? This also applies to this! So, after everything’s factored out, calculate the values of (or other variables) that will result the denominator to equal zero.

Now you’re done!

If you’re wondering ‘couldn’t we have just cancel numbers out before we factor’? Take note that if numbers have +/- between them, they’re called terms and you cannot cancel them. So we factor them so we can get factors, which has multiplication sign between them.

YOU CAN NOT CANCEL TERMS

YOU CAN ONLY CANCEL FACTORS!

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# Week 14 in Pre-Calc 11

One thing I’ve learned this week is about reciprocal functions. We’ve always use that word when we divided fractions. We ‘reciprocate’ the fraction we’re dividing with.

Anyways, let’s recall what a reciprocal is:

• the reciprocal ofis $\frac {1}{x}$
• Basically, the reciprocal of a number is just the numerator and denominator switched.
• Remember that all whole numbers and variables are over one. They have a denominator of one, basically.

What we need to remember about this topic are asymptotes, one’s called vertical asymptote (x value), one’s horizontal asymptote (y value).  An asymptote is a line that corresponds to the zeroes of your equation.

So, let’s say we have $y = \frac{1}{3x + 6}$

We must know that we cannot have a zero on the denominator, so we’ll first find out which value of x would result to have a zero denominator.

3x + 6 = 0

3x = -6

x = -2

Like was stated earlier, asymptotes are the zeroes to the function. So basically, since we just found out the zero of the denominator, we also found out one of the asymptotes. Remember, the vertical asymptote is a value of x, so our vertical asymptote is (x=-2)

Right now, though, we don’t need to bother with the horizontal asymptote. Take note that if the function’s denominator is 1, our horizontal asymptote will always be zero. (y=0)

So this is how you graph it:

• put a dashed line on where x= -2 is.
• put a dashed line on where y = 0 is.
• If the slope of your original line (3x + 6) is positive, then you will draw the hyperbolas on quadrant I and III, and if it’s negative draw it on quadrant II and IV.