Math 11 Week #8

Posted on October 29, 2018 by mollym2016 under Grade 11, Math 11, Pre-calc 11

This week, we learned all about parabolas. I had no idea what that word meant until we learned it, but now I have a pretty good understanding.

When graphing quadratic equations, the shape that comes up on the graph is called a parabola. A parabola is basically a U shape, with a vertex (the point at the very bottom or top). It is basically a bunch of repeating numbers, mirroring each other.

When graphing a quadratic equation, there are a lot of things we need to find :

Vertex:

x^2 + 3x + 6

For example the vertex on this graph is (-1.5,3.75) because that is where the parabola stops

Now, there can either be a maximum or minimum vertex, depending on the coefficient in front of the $x^2$ . If the coefficient is positive, then it is a minimum vertex because the parabola opens up (vertex = lowest point). If the coefficient is negative, then it is a maximum vertex because the parabola opens down (vertex = highest point).

In this graph, it is a minimum vertex because the U shape is opening up.

Line of symmetry :
The line of symmetry is the line that divides the two sides of the parabola apart, it is the value of x in the vertex. The example above’s line of symmetry is -1.5

X-intercept and Y-intercept:

The intercept is the numbers where the parabola crosses either the x or y value. Above, there is no x intercept but the y intercept = 6 because that is where the line crosses.

Domain and Range:
Domain and Range tell us what numbers fit into or work with this parabola and which don’t. Domain is the x values and Range is the y values. With our example, x can be all real numbers while y > or equal to 3.75 because there are no y values lower than that number

Math 11 Week #7

Posted on October 19, 2018 by mollym2016 under Grade 11, Math 11, Pre-calc 11

This week in math 11, we learned about the role of the discriminant in quadratic equations.

The discriminant is the radicand in the quadratic formula:

$b^2 - 4ac$

The discriminant tells us the nature of the roots of the quadratic equation, which basically means it tells us how many roots there are, if they are rational or not, and if they are real or not.

$b^2 - 4ac > 0$ – there will be 2 distinct roots

$b^2 - 4ac < 0$ – there are no real roots/unreal roots

$b^2 - 4ac = 0$ – there is 2 equal roots/one number

$b^2 - 4ac$ = perfect square- it is rational

$b^2 - 4ac$ = non-perfect square- it is irrational

For an example we are going to use a basic quadratic trinomial and find it’s nature of roots

$x^2-4x+5=0$

Now we find the discriminant with the formula $b^2 - 4ac$

(a=1, b=-4, c=5)

$(-4)^2 - 4(1)(5)$

$16 - 20$

$-4 < 0$

So we know that since the discriminant is less than 0, there are no real roots and that the root is unreal.

Math 11 Week #6

Posted on October 14, 2018 by mollym2016 under Grade 11, Math 11, Pre-calc 11

This week in pre-calculus, we learned 3 methods on how to solve a quadratic equation (an equation with a degree of 2).

My preferred way is called completing the square.

The example equation we are going to use is $x^2 + 6x +5 = 0$ .

So first off we need to make sure that this equation = 0, which it does already.

Next we need to write out the equation like so:

$x^2 + 6x$ + ____ – ____ $+ 5 = 0$ .

Now, to find the 2 numbers (which have to be the same), we take the b number which in this case is 6, we divide it in 2 and then square it.

$(\frac{6}{2})^2 = 9$

Now we insert this number into the blank spaces (9).

$x^2 + 6x + 9 - 9 + 5 = 0$ .

Next we use the first three terms and treat it like a trinomial and factor it, since it is pretty easy to do. In this case we would have to find a 2 numbers that multiply to 9 since it is the c term, but add up to 6 since it is the b term. In this situation, the number is +3.

$(x +3)^2 - 9 + 5 = 0$

Next we evaluate the last two numbers, -9 +5.

$(x +3)^2 - 4 = 0$

Next we need to solve for x, so it is now just easy solving. First we remove the -4 and move it to the other side and become +4.

$(x +3)^2 = 4$

Now, since we eventually need to isolate the variable, we need to square both sides.

$\sqrt{(x +3)^2} = \sqrt{4}$

$x +3 = 2$

Now, we isolate x by removing the negative 3 from the left side and making it positive three on the other.

$x = 5$

Now we have our solution!

Math 11 Week #5

Posted on October 7, 2018 by mollym2016 under Grade 11, Math 11, Pre-calc 11

This week, we reviewed factoring. I haven’t always been the best at factoring in grade 10 so it was a little tough for me to do the review because of how I couldn’t remember my method and also, I wasn’t the best at it in the first place.

I learned something new this week which made factoring less scary and difficult for me: the box method. It’s basically just a visual way to factor a trinomial.

Let’s use $3n^{2} + 2n -1$ as an example

So first you would insert the first term ( $3n^{2}$ ) and the last term ( $-1$ ) into the first and last spaces in a four square box

Next, we need to find the product of these two terms which ALSO is the sum of the middle term of the equation ( $2n$ ). What I do is just make a list like this: