This week in precalc 11 we learned about graphing quadratic inequalities which is in the form of y > a + bx + c.
(the sign could be <,>, ≤ , ≥)
We also learned how to graph linear inequalities in two variables.
First we need to rearrange the equation into the y = mx + b form. Then we use the y intercept and the slope to graph the points, we then connect these points into a straight line. depending whether the symbol is <> we use a broken line and if its ≥ or ≤ we use a solid line.
Finally we just color the side with the solutions, to do this we just pick a test point and use it on the original equation (usually 0,0 is a good test point), if the statement is true we shade in the side of the graph with the points if not we shade the other side.
This week in precalc 11 we had our chapter 4 unit test and our midterm. Some things I wish I studied a little more for the midterm was unit 2 and 3 where I know I was a little confused.
Then on Friday we had our first lesson on unit 5, where we were introduced to solving and graphing quadratic inequalities with one variable. With inequalities, we just solve the problem like a quadratic equation. First we factor, then we need to find the zeros of the equation, and then finally we need to test the numbers to make sure the statement is right.
Also one thing we were told to remember was that when we divide a negative inequality, you have to switch the sign.
This week in precalc 11 we learned about equivalent forms of a quadratic function. This just means finding different ways or representing a quadratic function to find more information about it. So we basically have
1.) Standard (Vertex) Form: y = a + q
We can find the vertex
2.) General Form: y = a + bx + c
You can find the Y-intercepts
3.) Factored Form: y = a ( x – x1 ) (x – y2 )
You can find the X-intercepts
We also learned how parent functions transform . Here we have the vertex form y = a + q
The a in the equation will tell us if the parabola opens down or up, and if there’s a stretch or a compression. the p will tell us if there’s a horizontal translation and the q will tell us if there’s a vertical translation.
This week in precalc we learned how to analyze quadratic functions. We also learned how to use desmos to graph and see how parent functions transform. A quadratic function is written y = a + bx + c .
*Note the difference between equations and functions.*
Equations: you can solve and find roots.
Functions: you can graph, and find Y-intercepts and X-intercepts.
The parabola is the curve where any point is at an equal distance from. The vertex is the highest/lowest point, it may be a minimum or maximum point. The line of symmetry is the line that divides the parabola into mirror images.
When analyzing quadratic functions in a graph, we always need to look for the y-intercept, the x-intercept, the domain and range, parabola, vertex, line of symmetry, if it’s opening down or up, and if it’s congruent to the parent function.
How do we know how parent functions transform:
here’s the video and the link to the website that helped me understand how parent functions transform:
On week 7 we learned a little about the discriminant, and the formula used is – 4ac . It helps determining how many solutions and real roots an equation has. The discriminant is pretty much the number you get under the square root in the quadratic formula.
If the answer to the equation is more than 0 there are 2 solutions, two distinctive real roots.
If the answer to the equation is equal to 0, then there is 1 solution (2 equal solutions), one real root.
If the answer to the equation is less than 0, there are no solutions, no real roots.
1.) 2 solutions
2.) 1 solution or 2 equal solutions
3.) no solutions
This week in precalc 11 we learned how to use the zero product law, which just simply means that if ab = 0, then either a = 0 or b = 0, or both. Let’s say for example we have + x – 30 = 0. First we factor the left side, ( x + 6 ) (x – 5 ) = 0 , now we use the zero product law, being x + 6 = 0 , and x – 5 = 0. Now we just simply solve this like any other equation and isolate X, and the final answer is either x = – 6 or x = 5 (these are the two solutions of the equation)
We also learned how to use square roots to solve quadratic equations, also called completing the square. This usually happens when the equation doesn’t factor, but luckily there are 3 steps you can use to do so. First, you take the middle term, then you divide it in half and finally, you square it.
Let’s say for example, we have + 6x + __
First we identify the middle term, which is 6x. Then we divide it in half, giving us 3, and then you just square it, giving us 9. At the end, the whole equation looks like this:
+ 6x + 9
This week in precalc 11 we did some factoring review. There were 5 terms we learned:
C- common: a good example that helped me understand this was 15 – 5 = 5 ( 3 – 1 )
D- difference of squares (binomials): an example that helped me understand this was ( – 64 ) = ( x + 8 ) ( x – 8 ) *they are like conjugates.
P- patterns (trinomials) : this one is a bit harder but after some practice it becomes easier. First you need to identify each number with the formula + bx + c, then you need to write down all factors pairs for c, and then you need to find a factor pair that sums up to b, and finally you substitute factor pairs into two binomials
E- easy patterns: an example that helped me understand this one is + 12x + 35 = ( x + 5 ) ( x + 7 )
U- ugly patterns: all the ugly polynomials that seem hard or just ugly in general: 5 – 7x + 2. For these, we learned to solve them using the box method.
This week in precalc 11, we learned how to add and subtract radicals. The steps do to so are pretty easy. First, we simplify the radicals in the problem, and then we need to make sure the indices and the radicands is the same. If they are the same, you add or subtract the coefficients with the coefficients, and the radicands just stay the same.
Just remember though, if the radicands are not the same, you can’t add or subtract them together.
An example that helped me understand this was:
Simplify the terms:
7 + 5 – 3
Add the coefficients together, and leave the same coefficient:
= 7 + 5 – 3 = 9
All you have to remember is to always simplify the radicals, it will make your life way easier!
This week in precalc 11 we learned about the absolute value of a real number.
The absolute value is the principal square root of a number, it is the distance from a to 0 on the number line. The absolute value is always positive. The sign used to represent it is | |.
For example, let’s say we’re trying to find the absolute value of | -39 | = 39 because the distance between -39 is 39 (the number has to be positive). Another example would be | 24 | = 24 which means that the distance between 24 and 0 is 24.
The absolute value bars do not work like parenthesis or brackets. For example, if you want to simplify – | -25 |.
Because the absolute value of negative 25 is positive 25, we end up with negative positive 25, which at the end gives us negative 25.
– | – 25 | = – ( +25 ) = -25
This week in precalc 11 we learned about geometric series.
Geometric series is the sum of the terms of a geometric sequences, a series with a constant ratio between successive terms. For example, a geometric series would be 6 + 12 + 24 + 48 + . . . The common ratio is the ratio between two numbers in a geometric sequence. To determine the common ratio, you can just divide each number from the number preceding it in the sequence (formula: r = a(n) / a(n – 1) ).
We also learned how the formula for determining the sum of the first n terms in any geometric series using the formula:
Here’s an example on how to apply this
Find for the geometric series 80 + 60 + 45. . .
a = 80
r = 0.75
= 80 ( ( ) – 1) ÷ ( o.75 – 1)
Another example ↓
For the geometric series 3, 9, 27. . . 6561, determine how many terms it has and then calculate its sum.
r = 3
a = 3
= a (
6561 = 3 (
2187 = (
7 = n – 1
8 = n
= 3 ( – 1 ) ÷ ( 3 – 1)