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

## Precalc week 4

This week in Precalc i was absent for the most part of the week so i only had two days of learning.   The most significant thing i learned was adding and subtracting radicals and defining variables.  Having the same radicand is only way to add and subtract radicals.  if the radicand is not the same then you must simplify the radical to find a common factor. $2\sqrt{4x}$ $+4\sqrt{4x}=6 \sqrt{4x}$

x is greater than or equal to 0.

In lesson 2.4 we learned about dividing radicals.  You must rationalise the denominator by multiplying the top by the conjugate and then distribute and simplify.  Don’t forget to FOIL. $5 \sqrt{5} - \sqrt{7 / }$ $5 \sqrt{3=}$ $3\sqrt{15} - \sqrt{21 / } 15$

In lesson 2.5 I learned about solving Radical equations.  I learned how to isolate a square root sign and determine the root of the equation and whether or not it’s a real root.  You can verify your answer by plugging x into the equation and checking to see if it matches. $\sqrt{5x + 3} = \sqrt{3x +1}$ $2x = -2$ $x=-1$

It is not a real root