For this week’s blog post, I picked the three equations of a line: slope-intercept form, point-slope form, and general form. My goal was to make myself feel more comfortable and understand that these equations so I could use them in a lot scenarios.

1. Slope-Intercept Form

Formula: y=mx + b

• m is the slope (how steep the line is).
• b is the y-intercept (where the line crosses the y -axis).

Steps:

1. Make sure the equation looks like y =mx + b.
2. The number in front of x is the slope (m).
3. The constant (b) is the y-intercept.

2. Point-Slope Form

Formula: y – y1 = M(x – x1)

• m is the slope.
(x1, y1) is a point on the line.

Steps:

1. Find the slope and a point on the line.
2. Replace m, x1, and 1 in the formula.
3. Simplify if needed.

3. General Form

Ax + By + C = 0

• A, B, and C are whole numbers.
• A should be positive, and no fractions are allowed.

Steps:

1. Move all terms to one side of the equation so it equals 0.
2. If there are fractions, multiply through to get rid of them.
3. Make A positive.

Examples

1. Slope-Intercept Form:
   A line with slope 3 and y-intercept – 2: y = 3х – 2
2. Point-Slope Form:
   A line with slope 1/2 passing through (4, -1):
   y+1=1/2(x-4）
   Simplified:
   y = 1/2x – 3

This week, I learned about parallel and perpendicular lines in math. At first, I didn’t fully understand the topic, so I decided to review it more thoroughly for my blog post. Here’s a summary of what I learned

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1. Parallel Lines

• Two lines are parallel if they never meet, no matter how far they are extended.
• Example:
  • Line 1: y=2x+3
  • Line 2: y=2x−5
  • Both lines have the same slope (m=2), so they are parallel.

2. Perpendicular Lines

• Two lines are perpendicular if they intersect at a right angle (90∘).
• For perpendicular lines, the slopes are negative reciprocals of each other.
  • If m1=a/bm_1 = a/bm1​=a/b, then m2=−b/am_2 = -b/am2​=−b/a.
  • One slope must be positive, and the other must be negative.
• Example:
  • Line 1: y=2/3x+1
  • Line 2: y=−3/2x-4
  • Since m1=2/3 m2=−3/2, the lines are perpendicular.

This week, we learned about function notation, mapping notation, and relations. I’m focusing on function notation because it’s important for understanding other math topics. Here are the steps to follow, with two simple examples:

Steps for Function Notation:

1. Know the function name:
   The name is usually a letter like f, g, or h.
2. Find the input value (x):
   The input is the number you will substitute into the function.
3. Plug in the input:
   Replace x with the given number in the equation.
4. Solve:
   Do the math to find the answer.

Example 1:

Let g(x)=4x−7
Find g(3):

Step 1: The function name is g.
Step 2: The input is x=3
Step 3: Replace x with 3:

g(3)=4(3)−7

Step 4: Solve:

g(3)=12−7=5

Answer: g(3)=5

Example 2:

Let h(x)=x2+2x−1
Find h(−2):

Step 1: The function name is h
Step 2: The input is x=−2
Step 3: Replace x with -2:

h(−2)=(−2)2+2(−2)−1

Step 4: Solve:

h(−2)=4−4−1=−1

Answer: h(−2)=−1

The domain and range in mathematics describe the inputs and outputs of a function:

1. Domain:

The domain is the set of all possible inputs ($x$-values) for which the function is defined. It excludes any values that cause division by zero, negative square roots (in real numbers), or any other undefined operations.

2. Range:

The range is the set of all possible outputs ($y$-values) that the function can produce based on its domain.

Example 1: f(x)=x^2−4

Domain: You can square any real number, so:

Domain=(−∞,∞)

Range: The smallest value of x^2 is 0 (when x=0), so the smallest output of f(x) is:

Range=[−4,∞)

Example 2: f(x)=x+3

Domain: There are no restrictions on $x$, so:

Domain=(−∞,∞)

Range: Adding 3 shifts all the outputs of $x$. Since $x$ can be any real number, the range is also:

Range=(−∞,∞)

Example 1: Finding the x- and y-intercepts for y=2x−6y

Finding the x-intercept:

1. To find the x-intercept, I set y=0 because the x-intercept is where the line crosses the x-axis (where $y$ is zero).
2. Substituting y=0 gives us the equation: 0=2x−6
3. Next, I solved for x by isolating it on one side of the equation: 2x=6 and x=3
4. So, the x-intercept is (3,0)

Finding the y-intercept:

1. To find the y-intercept, I set x=0 because the y-intercept is where the line crosses the y-axis (where x is zero).
2. Substituting x=0 gives us the equation: y=2(0)−6y 
3. So, the y-intercept is (0,−6)

For this equation, I learned that finding intercepts helps us sketch the line on a graph quickly since the intercepts provide two points through which the line passes.

Example 2: Finding the x- and y-intercepts for y=x^2−4

Finding the x-intercepts:

1. I set y=0 to find where the graph crosses the x-axis.
2. This gives us: 0=x2−4
3. To isolate $x$, I added 4 to both sides: x^2=4
4. Then, I took the square root of both sides, giving us two possible x-values: x=2 and x=−2
5. So, the x-intercepts are (2,0) and (−2,0).

Finding the y-intercept:

1. I set x=0 because the y-intercept occurs where $x$ is zero.
2. Substituting x=0 into the equation: y=(0)2−4
3. The y-intercept is (0,−4).

Prime Factorization is the process of breaking down a number into a product of prime numbers (numbers that have no factors other than 1 and themselves). A helpful way to find these prime factors is by using factor trees.

Steps for Prime Factorization

1. Start with the Number: Choose a number you want to factor.
2. Choose Factors: Think of two numbers that multiply to get that number. Write the number at the top, then draw two branches for the factors.
3. Repeat Until All Factors are Prime: For each composite (non-prime) factor, keep breaking it down into smaller factors. Stop when all branches lead to prime numbers.
4. List the Prime Factors: Once all factors are prime, list them.
5. Combine Like Factors with Exponents: Group any repeating prime factors using exponents to simplify the answer.

Example 1: Prime Factorization of 180

1. Start: Write 180.
2. First Split: Choose factors like $60$ and $3$.
3. Continue Factoring:
   • $3$ is already prime.
   • Break $60$ down into $30$ and $2$.
   • $2$ is prime.
   • Break 30 down into 1$5$ and $2$.
   • Break 15 into $5$ and $3$, which are both prime.
4. List Prime Factors: $2$, $2$, 3, $3$, and $5$.
5. Combine with Exponents: 180=2^2⋅3^2⋅5.

Example 2: Prime Factorization of 210

1. Start: Write $210$.
2. First Split: Choose factors like $2$ and 105.
3. Continue Factoring:
   • $2$ is prime.
   • Break 105 into $5$ and 21.
   • 5 is prime.
   • Break 21 into $3$ and $7$, both prime.
4. List Prime Factors: 2, 3, 5, and $7$.
5. Final Answer: 210=2⋅3⋅5⋅7

In class this week, we covered factoring polynomial. A polynomial can be factored by removing or dividing out the greatest common factor (GCF) from each term. The examples below show how we applied this method to different types of problems.

Steps
1. Combine like terms / simplify
2. Factor the resulting polynomial

in my First Ex, I combined like terms and then factored the result. This process involved simplifying the polynomial and finding the common factor.

in my Second Ex, I didn’t need to combine like terms because the expression was already simplified. I directly factored it by identifying the GCF.

Example 1:
5x^2−2x+7=2x^2+8x−7

3x^2+6x=3(x+2x)

Example 2:

9x^4 – 6x^3 + 3x^2

= 3x^2(3x^2 – 2x + 1)

This approach helps simplify complex expressions by breaking them down into more manageable factors.

This week in math class, I learned how to multiply binomials using the double distributive property. A binomial is an expression with two terms, such as (x+1) or (2x−4). To multiply two binomials, we distribute each term in the first binomial to each term in the second binomial, then combine like terms to arrive at the final answer.

Example 1

Problem: Multiply $(x+1)(x+3)$.

Steps:

1. Multiply $x$ by $(x+3)$:

   x(x+3)=x^2+3x

2. Multiply $1$ by (x+3):

   1(x+3)=x+3

3. Combine all the terms:

   x^2+3x+x+3=x^2+4x+3

so,  (x+1)(x+3) = $+3$.

Example 2

Problem: Multiply $(2x+5)(x-2)$.

Steps:

1. Multiply $2x$ by (x−2):

   2x(x−2)=2x^2−4x

2. Multiply $5$ by $(x-2)$:

   5(x−2)=5x−10

3. Combine all the terms:

   2x^2−4x+5x−10=2x^2+x−10

So, (2x+5)(x−2) = 2x^2+x−10

Multiplying binomials using the double distributive property involves distributing each term and combining like terms, which effectively simplifies expressions.

Mrs. Burton taught the class about “Exponent Laws,” and I learned three important ones: multiplication law, division law, and power law.

Multiplication Law:

When multiplying numbers with the same base, you add the exponents together. The base stays the same. For example:

3^2×3^4=3^2+4=3^6

If the bases are different, you multiply the bases normally, and the exponents stay as they are. For example:

2^3×5^3=(2×5)^3=10^3

Division Law:

When dividing numbers with the same base, you subtract the exponents. The base remains the same. For example:

7^5/7^2​=75^−2=7^3

If the bases are different, you divide them and keep the exponents. For example:

8^4/2^4​=(8/2​)^4=4^4

Power Law:

The power law applies when there’s a base with an exponent inside a bracket and another exponent outside the bracket. To solve this, multiply the exponents. For example:

(5^3)^2=5^3×2=5^6

Another example:

(2^4)^3=2^4×3=2^12