This week in pre-calculus, we learned about solving rational equations. Last week we learned how to solve expressions, but now I know how to find an actual solution for our variable. There are a couple ways to solve certain questions and in this post, I will be showing you my preferred ways: multiplying throughout by a common denominator and cross multiplying.

• Multiplying throughout by a common denominator:

First, we need to find the lowest common denominator. To do that we look at all of our denominators and multiply them together, without repeating a certain number/variable. In this case, we wouldn’t want to use both x’s because we already are using one.

For this equation, we can see that our lowest common denominator is 3x ((3)(x)). Now that we know that, we multiply every term by 3x.

If one of the terms in a denominator of a fraction is the same as the LCD, we can cancel that term out, making it easier to solve.

Now we just solve it like a simple linear equation that it is.

• Cross-multiplication:

If we are given a question where we have 2 fractions that equal each other, we can use the cross multiplication method.

With this equation, we multiply the numerator of the first fraction and the denominator of the second fraction and vice versa.

This week in math, we learned about reciprocal functions of parabolas. There are three forms of a reciprocal function graph for parabolas based on how many roots/where are the roots of a quadratic equation. They are similar to linear reciprocal functions, but a bit more complicated since there are can be up to two x-intercepts instead of just always having one with a linear function.

When we have a quadratic function with 2 x-intercepts like this: 

The reciprocal function looks like this: 

All values of x when y=1 and y=-1 are turned into the invariant points. The vertical asymptotes would be the original x-intercepts of the parent function (-5 and 1) and horizontal would be y=0.

When we have a quadratic function with no roots because it stays either below or above the x axis like this: The reciprocal function would look like this: 

Since there are no roots, there would be no vertical asymptote and the horizontal asymptote would be y=0. The invariant point would be the reciprocal of the y value of the vertex.

Finally, if you had a quadratic function with one root like:

The reciprocal function would look like this:

The invariant points would be where y=1 and and y=-1. The vertical asymptote would be the x value of the vertex and the horizontal asymptote would be y=0.

This week in Pre-calc 11 we started learning about how to graph absolute values. The parent function that we used is . As we know from our past units, an absolute value is the distance of a number from 0 on the number line, meaning that there will never be a negative y-value.

We used our past knowledge of graphing to help us better understand how to graph an absolute value. For example, we already know how to graph equations like (1) or (2)

1

2

With these to non-absolute value equations, we can see that there are y values that are negative. Now, we add in the absolute value symbols to the equation and see the difference from the two graphs.

(1) and (2)

1

2

After adding in the signs, we can see the difference it makes. For the first graph, the line stopped at the x axis and bounced off it, therefore reflecting itself on the opposite side. The point where the line stops at the x-axis and bounces back up is called the critical point (similar to the vertex of a parabola.

For the second graph, we can see that when no negative y values are allowed, the negative vertex and some other negative points had to flip to the positive side to create a W shape.

This week in pre-calc, we learned how to graph linear inequalities. In order to be an inequality, there would need to be >,<, ≥ or ≤ in an equation, displaying that the numbers on each side do not perfectly equal each other like they would if it was =.

When graphing, we use a dotted line or a solid line to express the differences in signs used in an inequality

> – greater than (dotted line)  – doesn’t include boundary line (the line itself)

< – less than (dotted line) – doesn’t include boundary line

≥ – greater than or equal to – does include boundary line

≤ – less than or equal to – does include boundary line

With linear inequalities, the inequality must have a degree of 1 (no visible exponents). Such as, y>2x-2.

If we use this as an example, graphing the line is simple since we already know how to graph a line because we learned it in grade nine.

Now, adding what we just learned, we need to find on which side of the graph that y is greater than 2x-2 (y>2x-2). In order to do that, we test points on our graph. If they turn out true, we know that the side the point was on includes all true points that satisfy the inequality.

The easiest point to graph is (0,0) – insert it into the y and x places in our inequality.

y>2x-2

0>2(0)-2

0>-2

0 is greater than -2, so this statement is true. With this information we know that the side of the graph that this point was on is the side with all possible solutions. To express that, we shade in that side, making sure to have a dotted line to show that the points on the boundary line are not included because of the >.

With that, our graph expresses our inequality and we are done!

This week in math, we started learning about graphing inequalities.

First, we looked at an inequality like . Since we already know about graphing normal equations that equal to 0, we know how to graph 

With that, we know how to find the x-intercepts of the inequality just because of our past knowledge, we use factored form. When factored, the inequality = (x+5)(x-2)>0 so we know that the x-intercepts are x=-5 and x=2.

We can know create a number line to express the 2 known points of this graph: 

Now, we need to find WHEN the graph is greater than 0 as the inequality expresses, so we can use test intervals. This is basically inserting a number as x from the three sections of our x-axis that we have created with our 2 points. We need a number less than -5, a number in between -5 and 2, and then the third will be greater than 2. We do these tests to see if the number that is given after the calculations makes the inequality true. If it is true, we know that is the part of the graph were the inequality is true.

Test intervals:
x = -10 (x<-5)

TRUE

x=0 (-5<x<2)

FALSE

x=10 (x>2)

TRUE

With this information, we can conclude that our solutions are that the inequality is true when x>2 and when x<-5

This week in math, we learned how there are three different forms or formulas for quadratic functions. Each one helps us discover an important part about the function and how to graph it.

1) $latex{y=ax^2 +bx + c$}$

General form

This equation easily gives us the y-intercept, as c = the y intercept.

For example if the equation was , we know the y intercept: (0,8)

2) 

Standard/vertex form

This equation gives us the vertex, with p = the x coordinate (the opposite) and q = the y coordinate in the vertex. V: (p,q).

Also, a = the stretch or compression of the function. and  When there is a stretch, it means the parabola will become narrower, and if there is a compression it will become wider. If a = 1, it is neither a stretch or compression, meaning the function is congruent to

For example, if we have the function , we know that the Vertex = (3,3). We also know that it is a stretch of 2.5.

In order to achieve this equation, we complete the square of the general form equation.

3)  y=a(x-x₁)(x-x₂)

Factored form

With this equation, we can find the 2 x-intercepts. x1 and x2 = x-intercepts.

For example, if we have the equation $latex y=2(x-7)(x-1), we know that the x-intercepts are x= 7 and x=1

To achieve this form, you need to factor the general form equation.

This week, we learned all about parabolas. I had no idea what that word meant until we learned it, but now I have a pretty good understanding.

When graphing quadratic equations, the shape that comes up on the graph is called a parabola. A parabola is basically a U shape, with a vertex (the point at the very bottom or top). It is basically a bunch of repeating numbers, mirroring each other.

When graphing a quadratic equation, there are a lot of things we need to find :

Vertex: 

x^2 + 3x + 6

For example the vertex on this graph is (-1.5,3.75) because that is where the parabola stops

Now, there can either be a maximum or minimum vertex, depending on the coefficient in front of the . If the coefficient is positive, then it is a minimum vertex because the parabola opens up (vertex = lowest point). If the coefficient is negative, then it is a maximum vertex because the parabola opens down (vertex = highest point).

In this graph, it is a minimum vertex because the U shape is opening up.

Line of symmetry :
The line of symmetry is the line that divides the two sides of the parabola apart, it is the value of x in the vertex. The example above’s line of symmetry is -1.5

X-intercept and Y-intercept:

The intercept is the numbers where the parabola crosses either the x or y value. Above, there is no x intercept but the y intercept = 6 because that is where the line crosses.

Domain and Range: 
Domain and Range tell us what numbers fit into or work with this parabola and which don’t. Domain is the x values and Range is the y values. With our example, x can be all real numbers while y > or equal to 3.75 because there are no y values lower than that number

This week in math 11, we learned about the role of the discriminant in quadratic equations.

The discriminant is the radicand in the quadratic formula:

The discriminant tells us the nature of the roots of the quadratic equation, which basically means it tells us how many roots there are, if they are rational or not, and if they are real or not.

– there will be 2 distinct roots

– there are no real roots/unreal roots

– there is 2 equal roots/one number

= perfect square- it is rational

= non-perfect square- it is irrational

For an example we are going to use a basic quadratic trinomial and find it’s nature of roots

Now we find the discriminant with the formula 

(a=1, b=-4, c=5)

So we know that since the discriminant is less than 0, there are no real roots and that the root is unreal.