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.