Solving QE's by factoring

Mathematics

1 answer:

gogolik [260]2 years ago

7 0

Answer:

The factored form of this is (x - 7)(x -4)

Step-by-step explanation:

In order to factor quadratic equaitons with a 1 as the first coefficient, we look for two numbers that multiply to the last number (28), but add up to the middle number (-11). To do this, we often list the factors of the last number.

Factors of 28

1, 28

-1, -28

2, 14

-2, -14

4, 7

-4, -7

The numbers -4 and -7 satisfy this requirement and are therefore the numbers we want. Now we stick them into parenthesis along with x terms in the front like this:

(x - 7)(x -4)

And that is the fully factored form.

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What are the points of discontinutity y=(x-3)/x^2-12x+27

madreJ [45]

Answer:

$(3, -\frac{1}{6})$

Step-by-step explanation:

We can rewrite the equation as

$y = \frac{x - 3}{(x - 3)(x - 9)}$

Notice that we have $x - 3$ in both the numerator and the denominator, so it looks like we can divide it out. However, what if $x - 3$ is $0$ ? Then we would have $y = \frac{0}{0 \times (x - 9)} = \frac{0}{0}$ , which is undefined. So although it looks like the numerator and denominator can be simplified, the resulting function we would get from simplification would not have the same behavior as this one (since such a function would be defined for $x = 3$ , but this one is not).

A point of discontinuity refers to a particular point which is included in the simplified function, but which is not included in the original one. In this case, the point which is not included in the unsimplified function is at $x = 3$ . In the simplified version of the function, if we plug in $x = 3$ , we get

$y = \frac{1}{((3) - 9)} = -\frac{1}{6}$

So the point $(3, -\frac{1}{6})$ is our only point of discontinuity.

It's also important to distinguish between specific points of discontinuity and vertical asymptotes. This function also has a vertical asymptote at $x = 9$ (since it causes the denominator to be 0), but the difference in behavior is that in the case of the asymptote, only the denominator becomes 0 for a specific value of $x$