Pre-Calculus 11 Week 8 – Graphing Quadratic Functions

April 9, 2018April 16, 2018 alexr2015

How do you graph a quadratic function? It’s the same way you would graph a normal equation. But instead of it looking like a line, it looks like this:

Like, literally, that’s it. That’s all. Just make a table of values, x and y on either side, plug some x values in, calculate for the y values. That’s all. But how bendy and twisty the line is isn’t random. You can spot certain patterns in quadratic functions. Take $y = z(x - p)^2 + q$ for example.

z represents the scale factor (More on that in a bit), and whether or not the parabola will open up or down. If it opens up, z is positive. If it’s negative, the parabola will open down.

p represents if the parabola moves to the left or to the right of the y-axis. If p is +5, then the parabola’s axis of symmetry, or the middle, will be x = 5. However, this means that in the equation, it’ll appear as $y = z(x - 5)^2 + q$ . Because if we remember, a positive and a negative make a negative, so in $y = z(x - 5)^2 + q$ , it’s actually telling us to translate five units to the right. If our equation was $y = z(x + 5)^2 + q$ (Notice the +), then the parabola would be translated 5 units to the left. p also represents the location of the vertex on the x-axis.

q simply shows us where the vertex lies on the y-axis.

As for z, z is usually 1. If the scale factor is one, the parabola follows a 1, 3, 5 pattern. As in one up, one over. Three up, one over. Five up, one over. And so forth. If the scale factor is 2, then all of that is multiplied by two. Two up, one over. Six up, one over. Ten up, one over. And so forth. Scale factor is 3? Three up, one over. Nine up, one over. Fifteen up, one over. And so forth. So the higher the scale factor is, the thinner the parabola is going to be. The lower, the fatter.

http://amsi.org.au/teacher_modules/Quadratic_Function.html

Pre-Calculus 11 Week 7 – Using the Discriminate to Help Solve Quadratic Equations

April 2, 2018April 2, 2018 alexr2015

This week, we learned how to learn if a quadratic equation had one solution, two solutions, or no solutions using the discriminate. What is the discriminate, you may ask? Look at your quadratic equation, see the expression inside the square root? That’s your discriminant. $b^2 - 4ac$

If you don’t remember, $a$ represents the first term’s coefficient, %latex b$ represents the second/middle term’s coefficient, and %latex c$ represents the third/last term’s coefficient.

Say your quadratic equation is $x^2 + 4x - 7 = 0$ , then the discriminate would be…

$(4)^2 - 4(1)(-7)$ $16 + 28$

The discriminant is $44$

The discriminant can tell us whether the answer has two solutions, one solution, or no solutions. If the discriminant is positive, then there are two solutions to your quadratic equation. If the discriminant equals zero, there will be one solution. However, if the discriminant is negative, then your quadratic equation has no solutions.

A quadratic equation graphed, showing two solutions.

What’s a solution? Glad you asked, if you attempt to graph a quadratic equation, your line isn’t a line! It’s this little bendy thing, which is called a parabola. For a quadratic equation to have a solution, a part of it has to be touching the x-axis. So, if the para

bola touches two points on the x-axis, then it has two solutions. However, if the apex of the parabola touches the x-axis, then that means the equation only has one solution. And if the parabola doesn’t touch the x-axis? No solutions.

A quadratic equation graphed, showing no solutions.

A quadratic equation graphed, showing one solution. Notice only the very top, the apex of the parabola touches the x-axis.

Pre-Calculus 11 Week 6 – Solving Quadratic Equations Using the Quadratic Formula

March 11, 2018March 12, 2018 alexr2015

This week, we’ve learned how to solve quadratic equations using the quadratic formula, which is different from simplifying When you attempt to solve a quadratic equation, there will be two possible answers. The quadratic formula looks like this (It’s not that scary, bear with me).

Now, in the quadratic formula, a represents the first term, b represents the middle term, and c represents the third term.

Say we’re given the question $3x^2 - 4x - 1 = 0$ . Here is what our quadratic equation, and our steps will look like:

You now have two answers for x, because you have a +or- sign.

Pre-Calculus 11 Week 5 – Factoring Quadratic Expressions

March 11, 2018March 12, 2018 alexr2015

This week, we learned how to solve quadratic equations. A quadratic equation looks something like this: $x^2 + 3x + 2$

$x^2 + x + n$ (n representing a number) Is what you’ll see most of the time. To solve these questions you factor it into two binomials. To check if your answer is correct, distribute the two binomials, and your answer should be the same as your quadratic equation.

The best way to find out what your two binomials are is to think about this: what are two numbers that multiply together to make your n term, but also add together to made your middle term? For the above question, your answer is simple. The only numbers that multiply together to make 2 are 1 and 2. If you add 1 and 2 together, you will get your middle term, 3! What about a harder question?

Our question is $x^2 + 7x + 12$ . I wrote down the multiples that could make 12. Then I find out which ones can add together to make our middle term, 7. The multiples 3 and 4 work! So our answer is (x+3)(x+4)!

Pre-Calculus 11 Week 4 – Adding and Subtracting Square Roots

February 26, 2018March 11, 2018 alexr2015

This week, we discussed adding and subtracting numbers with square roots, such as $2\sqrt{2} + 3\sqrt{2}$

The easiest way to learn how to add these numbers is simply looking at them the same way you would look at 2x + 3x in algebra. You can simplify this expression. 2x + 3x = 5x

But you wouldn’t be able to add, say, 2x + 2y together. The expression is already simplified. The same thing applies to roots!

You can add $2\sqrt{2} + 3\sqrt{2}$ together to make $5\sqrt{2}$ !

However, you are unable to simplify $4\sqrt{2} + 5\sqrt{3}$ , because the radicands aren’t the same.

If our square root is really big, like $\sqrt {128} + 3\sqrt{2}$ , we can still add them together! We just need to simplify the square root first. In this case, $\sqrt {128}$ simplifies into $8\sqrt {2}$ . Our steps are as follows:

$\sqrt {128} + 3\sqrt{2}$

$\sqrt {8*8*2} + 3\sqrt{2}$

$8\sqrt{2} + 3\sqrt{2}$

$11\sqrt{2}$

All these rules apply to subtraction as well.

$7\sqrt {3} - 3\sqrt{3} = 4\sqrt{3}$

and

$\sqrt {243} - 3\sqrt{3}$

$9\sqrt {3} - 3\sqrt{3}$

$6\sqrt{3}$

Pre-Calculus 11 Week 3 – Absolute Value of a Real Number

February 19, 2018February 19, 2018 alexr2015

This week, we looked at what absolute values were. Absolute values are the distance that the number is from zero. In class, we used the analogy of attempting to throw crumpled paper balls into a garbage bin. One person made the shot, others missed and landed a certain distance from the trash can. While some paper balls will have landed in different positions, their distance from the trash can could be the same as another paper ball. Absolute values are written like so:

|-5| = 5

The answer is 5 because -5 is 5 away from 0.

|5| = 5

The answer is also 5 because 5 is 5 away from -5.

Numbers beside these terms act as coefficients. Like so:

3|-5|

3×5

=15

Or if there is an expression within the lines, you do those first before completing the equation:

3|-3+6|

3|3|

3*3

= 9

You can almost think of the lines like brackets.

Pre-Calculus 11 Week 2 – Geometric Sequences

February 12, 2018February 13, 2018 alexr2015

This week, I focused on geometric sequences.

Geometric sequences share similar characteristics to arithmetic sequences. For one, geometric sequences are patterns, just like our arithmetic sequences from last week! Arithmetic sequences have common differences, such as this one: “5, 10, 15, 20, 25, 30…” You can tell the pattern here is that 5 is being added for every new term!

Geometric sequences, however, have common ratios. Like so: “2, 4, 8, 16, 32, 64…” You should be able to see a pattern here. What’s the pattern? The terms are being multiplied by 2 each time! This is very different from our arithmetic sequence, where numbers were being added as the pattern was continuing. The ratio is determined like so: tn/(tn-1)=r.

The ratio is a very important number. It’s as important to geometric sequences as the common difference, d, is to arithmetic sequences, so get used to finding it!

There are a number of ways that you can utilize the common ratio, r. Say, you wanna find the 9th term in our geometric sequence here. Looks something like this:

t1*r^(n-1) = t9

2*2^(9-1) = t9

2*2^8 = t9

2*256 = t9

512 = t9

The general formula for geometric sequences is t1*r(n-1)=tn

Pre-Calculus 11 Week 1 – Arithmetic Sequences

February 12, 2018February 13, 2018 alexr2015

First week of Pre-Calc 11! This week, I learned about Arithmetic Sequences.

Arithmetic sequences are like patterns, they look something like this: 5, 7, 9, 11, 13….

For a pattern to be an arithmetic sequence, all the numbers must have a common difference between them all. The common difference is how much it goes up or down. For our pattern here, the common difference is 2! You see how the pattern continues to increase by 2? This is an arithmetic sequence.

If you wanted to find the 20th term in that pattern, you would use the following formula:

$t_10=t_1+9d$

$t_10 = 5 + 9(2)$

$t_10 = 5 + 18$

$t_10 = 23$

With this formula, we’re saying that we’re going to take the first term in the arithmetic sequence, add the common difference (2) NINE times to get the 10th term. Let’s count it, and make sure my work is correct.

5, 7, 9, 11, 13, 15, 17, 19, 21, 23

Look at that, the 10th term really is 23! For this formula, you don’t have to use t1. You can use any term you already have. If we only had t4, the formula would look like this:

$t_10=t4+6d$

$23 = 11 + 6(2)$

$23= 11 + 12$

$23 = 23$

You can find a general formula for tn as well. It looks like this” $latex t_n = t1 + (n-1)d For example:

$t_23 = 5 + (23-1)d$

$t_23 = 5 + (22)2$

$t_23 = 5 + 44$

$t_23 = 49$

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