## Precalc 11 – “Week 17”

This week in Precalculus 11 we learned about trigonometry. During the week we improved our understandings of trig and learned two new laws for trigonometry.

The first law is the Sine Law:

$\frac{a}{sinA}=\frac{b}{sinB}=\frac{c}{sinC}$

The Sine law is used for triangles with no angle being 90 degrees. It can only be used when A, a and either another side or another angle; A, B and c.

The second law is the Cosine Law:

$a^{2}=b^{2}+c^{2}-2bcCosA$

The Cosine Law can only be used for triangles that have the three sides or when a,c and B are known.

## Precalc 11 – “Week 15”

$\frac{4}{x-5}+\frac{x-7}{5-x}$

Because $x-5=-1(5-x)$, Therefore

$\frac{4}{x-5}+\frac{-x+7}{x-5}$

$\frac{4-x+7}{x-5}$

$\frac{11-x}{x-5}$

## Precalc 11 – “Week 14”

This week in Precalculus 11, we learned about graphing reciprocal functions. The hardest part about this was graphing quadratic reciprocal functions, because there are three different forms depending on how many roots it has.

Here are the three different ways:

1.Contains one roots:

$y=\frac{1}{x^2+6x+9}$

2.Contains two root:

$y=\frac{1}{x^2+6x-9}$

3.Contains zero roots:

$y=\frac{1}{-x^2+10x-30}$

## Precalc 11 – “Week 13”

This week in Precalculus 11 we learned about Absolute Value Functions. We have already done absolute values which is:

$\mid-3x+5\mid=11$

Which is $3x=6$

$x=2$

Absolute values turn the equation positive. Therefore an Absolute Value Function can only be positive.

$y=\mid-3x+5\mid$

As you can see the blue line is $y=-3x+5$, and because absolute values need to be positive then the line cannot be below the x-axis, therefore it is flipped as you can see by the red line: $y=\mid-3x+5\mid$

## Precalc 11 – “Week 12”

This week in Precalculus 11 we learned about how to solve quadratic systems.

There are multiple ways of find the solution for where both lines cross, like graphing or elimination. However in class we learned about the process of substitution.

Which is removing one variable from the equation and then solving for the remaining unknown variable. After that using that answer to find the first unknown variable.

ex.

$y=2x+5$

$y=x^{2}+6x+9$

$2x+5=x^{2}+6x+9$

$0=x^{2}+4x+4$

$x=2$

$y=9$

## Precalc 11 – “Week 11”

This week in precalculus 11, we learned about graphing linear inequalities.

This is what we learn:

If the inequality was greater than or equal to y:

$y=<2x+2$, then you also need to shade the line.

Same goes for an inequality which is less than or equal to y.

However if it is only greater than or only lower than, then you must leave a dotted line to inform that the line does not count as part of the answer.

## Precalc 11 – “Week 10”

This week in Pre-Calculus 11, we studied and reviewed for our midterms next week. The most challenging thing for me is how to find $t_{8}$ when you only know $t_{2}=10$ and $t_{4}=20$.

With the help of some of my classmates, we realized 4 divided by 2 and 20 divided by 10, would give us the d value. by dividing those two answers: 10 divided by 2, it gives the d value which is 5. Then using the arithmetic formula you can find $t_{1}=5$ and then $t_{8}=40$.

## Precalc 11- “Week 9”

Modelling quadratic equations is when you are not given an equation, but rather words/sentences to describe how the equation works. This can be seen in this example:

The addition of two numbers is 30, by multipling the numbers you can get an answer of 200. So using these two phrases you can create two equations to find the x and y value.

$x+y=30$

this can also be written as $y=30-x$

$xy=200$

Now using the two equations you can find the x and y value:

$(x)(30-x)=200$

$30x-x^{2}=200$

$-x^{2}-30x-200=0$

$x_{1}=20$, $x_{2}=10$

Voila you have found the x and y value.

## Precalc 11- “Week 8”

There are three equations you can get:

$y=(a+b)(a+c)$

$y=ax^2+bx+c$

$y=a(x-p)^2+q$

Each one tells us a different thing about the graph.

$y=(a+b)(a+c)$ :

from this equation we get the x-intercepts.

$y=ax^2+bx+c$ :

from this equation we get the direction of the parabola, the narrowness/wideness and the y-intercept.

$y=a(x-p)^2+q$ :

from this equation we get the direction of the parabola, the narrowness/wideness and the vertex.

## Precalc 11 – “Week 7”

This week in Pre-calculus 11 we learned about Discriminants.

The discriminant is the part under the square root:

$x=\frac{-b(+or-)\sqrt{b^{2}-4ac}}{2a}$

So

$b^{2}-4ac$

Now if $b^{2}-4ac>0$, then there are 2 real roots.

If $b^{2}-4ac=0$, then there is only 1 real root.

If $b^{2}-4ac<0$, then there is no real root.

These are the three possibilities for the discriminant.