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This week we learned how to add and subtract rational expressions. These are a bit harder than multiplying and dividing because the denominator has to be the same throughout the expression.

Steps to simplify:

1. determine lowest common denominator (LCD)
2. rewrite each fraction as an equivalent fraction with LCD
4. reduce if possible

For example: $\frac{x-9}{2x}-\frac{3x}{x-4}$ where “x” cannot equal 0 or 4.

1. For this expression, the least common denominator would be $2x(x-4)$ because neither denominator can be simplified any farther or multipled to create a common denominator, so we multiple the two denominators together to create one LCD.
2. Since we’re multiplying the bottom, we also have to do the same to the top, so the equation becomes: $\frac{(x-9)(x-4)}{2x(x-4)}-\frac{(3x)(2x)}{(x-4)(2x)}$.
3. This simplifies to $\frac{x^2-13x+36-6x^2}{2x(x-4)}$
4. The final simplified expression will be $\frac{-5x^2-13x+36}{2x(x-4)}$

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This week we learned how to graph quadratic reciprocal functions, and learned some new vocabulary.

quadratic reciprocal functions will look one of three ways when graphed:

1. the parent function only touches the x-axis in one spot
2. the parent function doesn’t touch the x-axis at all
3. the parent function crosses the x-axis in two places

When graphing the parabola, you first find the asymptotes, invariant points, and then the location of the hyperbola.

Asymptotes are located at where the parabola crosses the x-axis. At the x-intercept(s) you can drop a dotted line vertically, and a dotted line horizontally. Depending on how many x-intercepts you have will dictate how many asymptotes you have.

Dotted in red:

Invariant points are located where your parabola meets the points on the y-axis 1 and -1. The parabola can have 0-4 invariant points. These dictate where you’ll draw the hyperbola.

Circled in blue:

Your hyperbola is the graph of the reciprocal of your original equation. For example: $2x^2+4$ becomes $\frac{1}{2x^2+4}$

When you graph the hyperbola, there’s no need to graph each individual point. Instead, you can just draw a curved line going through your invariant point and approaching zero.

Colored in purple:

This week we learned how to solve systems algebraically with more than one variable.

For example: x+6y=4 and 0=x-4y

In these types of systems you used steps.

1. Isolate a variable
2. Plug the equation for that variable into the second system
3. solve for that variable
4. plug variable into original equation, solve for the missing variable
5. Check!

This may not make complete sense in steps, so I’ll demonstrate using the first example.

1. x+6y=4 can be rearranged to make x=4-6y
2. next you plus your new equation x=4-6y into your second equation: 0=x-4y which turns into 0=(4-6y)-4y. This becomes y=2/5.
3. Next, we plug our value for y into our original first equation x+6y=4, which becomes x+6(2/5)=4
4. Simplifying the equation: x=1.6
5. Check y=2/5 and x=1.6: x+6y=4 -> 1.6+6(2/5)=4 Yes! 0=x-4y -> 0=1.6-4(2/5) Yes! So we know we did it correctly

While solving quadratic equations, it’s useful to take a look at the discriminant.

The discriminant is area under the square root in the quadratic formula:

if you have a quadratic equation (equation equal to zero with 3 distinct parts), you can use the quadratic formula to solve. Depending on the answer, we can figure out whether the equation will have 1,2 or 0 solutions.

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Let’s find the discriminant of the equation: $x^2-3x+6=0$

In a quadratic equation, the parent would be $ax^2+bx+c$

following the parent, in our equation, a=1, b=-3, and c=6

using this information we can plug in the numbers to our equation to find the discriminant.

If $b^2-4ac$ is our equation, we just put our numbers we found in the spots of the letters, and simplify.

$b^2-4ac$

$-3^2-4(1)(6)$

$9-4(6)$

$9-24$

$-15$

using this we know that since the discriminant is -15, the original quadratic equation does not have any solutions, and using the quadratic formula would not work.

I’m talking like that on purpose. Sorry if i offend anyone!