This week in pre-calculus 11 I learned about modeling problems with quadratic function. I was a little bit confused about how to solve that kind of questions but after few practicing, I learned how to solve those questions.
Here is an example:
This week in precalculus11 I learned about analyzing the quadratic function of the form . Each of these words represents something on the graph.
a is the coefficient. It shows the stretch or compression of the parabola. If a > 0 parabola becomes skinnier, and if a < 1the parabola becomes wider. Also, a shows whether the parabola opens up or reflect down.
P represents horizontal translation.
q represents vertical translation.
Here is an example:
This week in precalculus 11 I learned about discriminant. It was an easy lesson. The expression , is called discriminant of the quadratic equation because it discriminates among the types of possible solution. Here is an example:
calculate the value of the discriminant of this quadratic equation.
first step: write down the equation first
second step: write down what is a, b and c
a= 2 b=-8 c= 4
third step: put everything in the equation
64+32=96
this equation has two real roots so 97> 0
This week in Pre calculus11, I learned about the geometric sequence. In this lesson, I had a little bit of struggling when the question gives me two terms and asks for the unknown term. After understanding what the question is asking I learned how to solve this type of questions. Here is an example:
This week in Pre calculus11, I learned about the Arithmetic sequence. This lesson was an easy lesson, but I had a little bit of struggling at first when I had to solve the question that seemed a little bit hard for me, but after several practicing, I learned how to solve the question, Here is the Example:
Example 1: How many terms are in the sequence 2, 6, 10, 14,…….,50
first step:
we know the first term is 2, the common difference is 4, and we say the is 50
second step:
the equation is=> = +(n-1)d
third step:
Now we could put the things that we know in the equation:
50= 2 +(n-1)4
forth step:
we don’t know what is n in (n-1) so we would multiply the common difference which is 4 by n and -1
50= 2 +4n-4
Last step:
so now we put the unknown on one side and the known on the other side
50-2+4= 4n
52= 4n
=
n= 13
so there are 13 terms in that sequence