An Absolute value (||) means whatever inside is going to be positive, when we put a linear function within ||, it becomes an absolute value linear function.
The graph of it will change, as we know, a linear function is a straight line, either having a 0 slope or / positive slope or \ slope.
When graphing an absolute value linear function, if the slope is not 0, the line will bounce back up in the opposite direction of the original function at x-intercept.
Why is that?
Let me show you an example:
2x+4=y, a simple straight line with a slope of 2, and y-intercept of 4.
But magic happens when you put||on to the function. As the output of an absolute value cannot be negative, y, in this case, will also have to positive. But with our previous graph of the original equation, the Y clearly dipped under 0.
Let’s see what happens if we put an x value that would make y<0 into the absolute value equation. |2x(-4)+4|=4,
the Line reflected up, from this example we can learn that when graphing a linear absolute value function, all you need to do is find the x-intercept of the original line and draw a reflected line from the original root.
In previous grades, we learned about linear equations and inequalities with one and two variables, now we learn how to solve them graphically.
To graph a linear inequality with two variables we firstly rearrange it to y-intercept form in order to graph it.
With y-intercept form of the equation, we are able to graph a line just how we did in grade 9.
The difference is in an equality the equation represent the values on the line and inequality represent a whole area.
A line will divide the graph into two zones, we just need to determine which zone to shade(it represent a range of value)
The most accurate way of determining such things is by trial and error,
we would take two test coordinates from each section of the graph and plug it into the equation, if the equation is true then the side with the coordinates will be shaded.
EX: (0,0) , (-5,5)
2(0)+6<4(0) 6 is not less than 4, therefore the otherside will be shaded.
One last thing you have to do before you submit your answer is to determine whether you should use a solid line or dotted line if the sign involves = sign, you use a solid line.(In this case it’a dotted line due to the fact that it’s only a < sign).
In week 9 of pre-cal 11 We learned how to analyze the general form of quadratic equation along with many things.
With analyzing, I mean you factor the equation if it’s factorable. And apply the knowledge I learned from last unit to find the two or 1 roots, and using logical thinking such as the axis of symmetry must have the same distance on the x axis to both roots and etc to find more information without changing the equation into standard form.
In week 8 of pre-cal 11, we learned how to graph quadratic equations with the general form and the standard form.
a+bx+c, is the general form of the equation, with this form the equation we can find limited information comes to graphing; we can determine the direction and the compression ratio of the graph from a, and y-intercept from c.
But we can easily change this equation to standard form by completing the square; from this form of the equation we can determine the position of the vertex with q being the y-axis(+ going up,- going down), p being the x-axis(- going right, +going left) and the direction of opening and the compression ratio of the graph by looking at a,( the ratio of the graph will be congruent to a=y, a determines the direction of the graph( if negative a, facing down, if postive a opens up).
During week 7 of pre-cal 11, we learned how to find the discriminant of a general form of a quadratic equation.
In this quadratic formula:
The discriminant is the part under the root sign:
Finding the solution to the discriminant of a quadratic equation will provide you will the type of roots you will get from the aforementioned quadratic equation.
To find the discriminant of a quadratic equation you need to learn the meaning of a,b,c in this case:
a+bx+c, is the general form of the equation, and when you are trying to find the discriminant just put -4ac in, such as:
4+2x+6, the discriminant will be -4(4×6)=-92
And the range the discriminant is in it will provide you with the number of roots you will get from this quadratic equation; If your answer is a positive number, there are 2 roots, If your answer is a negative number, there are no real roots, If your answer is a 0, there is 1 root.
This week we learned a universal and fail-safe method of solving a quadratic equation, that is the quadratic formula:
You might be wondering what do all the letters represent, well in any quadratic equations you will be able to rearrange it into the form of ax2+bx+c then you will be able to solve this equation with the quadratic formula. This is the easiest and the safest method of solving a quadratic equation in my personal opinion.
This week we reviewed some of the basics of factoring from math 10 pre-cal, and we learned a phrase which will help us in using the optimized steps to determine the method of factoring that can be used for the expression. There are three big types of expression that we can factor which we are exposed to, the easy ones, the harder ones and the difference of Squares.
This week we reviewed the factoring of the binomials and the trinomials. In the case of the trinomials, a factorable expression must be written in the form of x2, x, n, there are two types of trinomial expression the easy ones: which the number before the term x2 is one, and the hard ones: the number before the term x2 is not 1. The methods of factoring these expressions use the first and the last term of the original expression as they are the result of only two number multiplying, and there are different method from there depends on the situations.
And there is the difference of squares which is arguably the easiest one, it is written in a binomial format with 2 square terms: x2- some square number. The method of factoring difference of square is extremely easy, you can find a pair of the conjugate with the second term of the conjugate equals the square root of the second term of the original expression and leave the first term as x.
This week we learned the calculations that involve radicals such as add, subtract. In the case of add and subtract we need to convert the subjects that we are adding into mixed radicals with the same radican. EX: when the radicals have the same radican you will have to add the exponent of the mixed radical and keep the radican the same. For multiplication and division you can just apply the rules of multiplication and division on entire radicals with one exception, that is you can’t write denominator as a radical.
In this week’s classes, we learned the concept of absolute value…
Absolute value expresses the distance between numbers on a number line, due to the fact that there can’t be a negative distance as it is a scalar value, everything coming out of the “||” are positive. and the expression within absolute value get prioritization in an equation, such as