## week 12 – Precalc

This week in Precalc we continued with unit 8 Absolute value functions.

We started the week with Absolute value and reciprocal functions.  The parent function is (y=x) to base all other functions off from.  If we have an equation like $|3x+6|$ we can use this to find the critical point (the point where the line bounces up) or the x-intercept which will be (-2).  from there we can find other features like the y-intercept (0,6), the domain ( x is all real numbers) and the range ($y \textgreater (or equal to) 0$.

The next thing we learned was absolute value expressions.  By simplifying, factoring, solving etc. we can find the solutions to the equations which are the solutions.

Ex :

$3=|2x+5|$

$2x+2=0$

$2(x+1)=0$

$x=-1$

$-2x-8=)$

$-2(x+4)=0$

$x=-4$

solutions are -4 and -1.

## Precalc week 11

This week in precalc we continued with graphing inequalities and quadratic functions.

In lesson 5.3 we learned how to graph quadratic functions with two variables instead of one.  the process is almost the same as one variable except we are dealing with parabolas now.

Ex :

$y \textgreater x^2+6x+10$

$= (x+3) -1$

Now using this information we can graph a parabola using the vertex (-3, 1) and the a value of 1.  An extension to this is now we have a inequality so that means we have to shade in part of the graph.  WE know that the parabola is facing up and the inequality sign is less than so that will mean we will have to shade in outside of the parabola.  The line creating the parabola will also be a solid line.

In lesson 5.4 we learned about intersecting inequalities on a graph and how to solve them.  The point where both parabolas or lines are touching each other marks the point of intersection.

Ex :

$y=x^2+5$

$y=-x^2+7$

After graphing these two parabolas, the points of intersection are (1,6) and (-1,6) which is the answer.  Sometimes there can be an infinite amount of points and sometimes there cannot be any points of intersection.

## week 10 – Precalc 11

This week in Precalc we had our midterm Wednesday.  We started unit 5 with solving inequalities with one variable (x).  We found that when solving for x we can use the answer to create an inequality based on the sign of inequality.

Ex :

$4x^2+8x+3 \textgreater 0$

$(4x^2+6x)(2x+3) \textgreater 0$

$(2x+1)(2x+3)$

$x= \frac{-1}{2} and \frac{-3}{2}$

We can use a number line or graph to write solutions for the quadratic function.

$x \textless \frac{-3}{2}$

$x \textgreater \frac{-1}{2}$

## Week 9 Precalc 11

This week in Precalc 11 we finished unit 4 and started to review for the midterm neat the end of the week.

In lesson 4.6 we discovered Equivalent Quadratic equations.  When you can change functions to different forms such as vertex form, general form and factored form.

General form to standard form = complete the square

factored form to general form = FOIL

general form to factor form = factor

vertex form to general form = expand and simplify

In lesson 4.7 we learned about modelling problems with quadratic equations.  this was the hardest lesson of them all since it was all word problems and you have to be careful of what the question is asking for.

Ex :

Two numbers have a difference of 18.  the sum of their squares is a minimum.  determine the numbers.

$a-b=18$

$latex.a^2+b^2=minimum$

$a=18+b$

$(18+b)^2+b^2$

$2b^2+36b+324$

$2(b+9)^2-162+324$

$2(b+9)^2+162$

$b=a+18$

$b=-9+18$

$b= 9$

the numbers are -9 and 9.

## Precalc week 8

This week in Precalc we continued with quadratic functions and analysing them to be able to graph them and how to use information given to interpret the equation to be able to graph it.

We started the week with a skills check.  When trying to graph an equation, you can look for clue in the equation that will indicate what quadrant the parabola is located in.  I learned that the function $y=x^2-5$ means that the parabola will be below 0 so near the bottom of the graph opening up.  When the function is $y=(x-3)^2$ it means the parabola will be in quadrant 1 and moved 3 units to the right.

In lesson 4.4, when learned how to analyse $y=a(x-p)^2+q$.  If $|a| \textgreater 1$, then it’s a stretch.  If $|a| \textless 1$ then its compressing.

We also learned that if you complete the square of a function, you will find the vertex.

Ex :

$y=x^2+8x+2$

$y=(x+4)^2-14$

Vertex = (-4, -14)

We also know that if provided with certain information about a graph, we can determine a variable such as A.

Ex :

Vertex : (-5, 2)

Y-int : -8

$-8=a(0+5)^2+2$

$-8 =25a+2$

$\frac{-2}{5} = a$

$y = \frac{-2}{5}(x+5)^2+2$

## week 7 precalc 11

This week in Precalc 11 we finished unit 3 with the quadratic equation and interpreting the discriminant.

In lesson 3.4 we learned how to use the quadratic formula : $x=-b \mp \frac{\sqrt{b^2-4ac}}{2a}$ to solve quadratic equations.  i found this especially useful with fractions and decimals.

Ex :

$x=2x^2-6x+1=0$

$x=\frac{6 \mp \sqrt{6^2-4(2)(1)}}{4}$

$x=\frac{6 \mp \sqrt {28}}{4}$

$x=\frac{3 \mp \sqrt{7}}{2}$

In lesson 3.5 we learned how to interpret the discriminant : $b^2-4ac$ and how to find whether they are equal roots, real roots, no real roots etc.  We also learned how to find the value of a variable when the equation has no real roots, one real root etc.

$b^2-4ac \textgreater 0 = 2$ solutions/real root

$b^2-4ac = 1$ solution/equal root

$b^2-4ac \textless 0 = 0$ solutions/no real roots

Ex :  $8x^2-5x+k=0$

determine the value of K when the equation has no real roots:

$5^2-4(8)k \textless k$

$25 - 32k \textless 0$

$\frac{25}{32} \textless 32k$

$k \textgreater \frac{25}{32}$

## week 6 – precalc 11

This week we started a new unit of solving quadratic equations.  last week was mostly a review for the harder lessons we had this week.  in lesson 3.2 we started by solving quadratic equations by using the zero product law.

$x^2-6x+5=0$

$(x-1)(x-5)$

$x= 1, 5$

Verify :

$5^2-6(5)+5=0$

In lesson 3.3 we learned how to solve perfect square trinomials by completing the square.  remembering to put a plus and a minus symbol when adding a radical is an important step when solving and verifying.

$x^2+4x=2$

\$latex x^2+4x-2=0

$(x+2)^2-6=0$

$(x+2)^2=6$

$x=-2 \pm \sqrt{6}$

## Precalc week 5

This week in Precalc we finished unit 3 and started to review factoring for the upcoming unit, “quadratic equations”.  In lesson 3.1 we reviewed how to factor easy trinomials, difference of squares trinomials, hard trinomials and perfect square trinomials.  these are all important things to know for the upcoming units.

difference of squares:

$x^2-100$ $=(x-10)(x+10)$

easy trinomials :

$m^2+4m-45$ $=(m-5)(m+9)$

hard trinomials :

$5x^2+9x+4$ $=(5x+4)(x+1)$

perfect squares :

$9x^2+6x+1$ $=(3x+1)^2$