xy-coordinate Plane
- Origin: (0,0) where the two axes intersect
- Quadrant 1 = (+, +)
- Quadrant 2 = (-, +)
- Quadrant 3 = (-, -)
- Quadrant 4 = (+, -)
- x-intercept = point where a line intersects the x-axis
- y-intercept = point where a line intersects the y-axis
- Line segment = line that terminates, has two distinct endpoints
- Line = line that is non-terminating, goes on forever
- Parabola: A plane curve which is mirror-symmetrical and is U-shaped
- Hyperbola : Looks like two parabolas opening in opposite directions.
Reflections and Symmetry across axis
- Reflection about/across the x-axis: make y neg (x,y)→(x,-y)
- Reflection about/across the y-axis: make x neg (x,y)→(-x,y)
- Reflection about/across the origin: make both x and y neg (x,y)→(-x,-y)
Distance Between Two Points
What is the distance between points (-5, 3) and (-3, -8)?
For points (x₁, y₁), (x₂, y₂)
Distance = √[(x₂ - x₁)² + (y₂ - y₁)²]
Distance = √[(-3 - -5)² + (-8 - 3)²]→√[(2)² + (-11)²]→√[4 + 121]
→√125 = 11.2
Linear Equations on the Plane
Linear equations (1st degree equations with one or more variables or constants) will consist of at most two variables: x and y. However, it’s also possible to have a linear equation consisting of only one variable
- One-variable equations:
y=constant : Horizontal line (y=4, every point on the line is 4 points above the x-axis)
x=constant: Vertical line (x=2, every point on the line is 2 points above the y-axis) - Two-variable equations: Will appear as a “slanted,” or “sloped,” line
y = x -2
Lines and Intercepts
- x and y intercepts of a line through the origin is (0,0)
- A horizontal line (y=3) has a y-intercept but no x-intercept
- A vertical line (x=3) has a x-intercept but no y-intercept
- 𝑥-intercept x y intercept = negative→slope is positive.
- 𝑥-intercept x y intercept = positive→the slope is negative.
- To find the x-intercept of an equation, set y = 0
In the case of y = (x - 3)(x + 4), x-intercepts are 3 and -4 - To find the y-intercept of an equation, set x = 0
Calculating Slope
What is the slope of a line containing the points (3, -5) and (-2, 7)?
- To calculate the slope of a linear equation (sloped/slanted line) If you have two points on that line, you can calculate the slope using the formula below, assuming the two points are (x₁, y₁), (x₂, y₂)
Slope Formula = (y₂ -y₁) ÷ (x₂ - x₁)
- Line rising left to right: Positive slope
- Line declining left to right: Negative slope
- Horizontal line: the slope is 0
- Vertical line: the slope is “undefined.”
(7 - -5) ÷ (-2 - 3) = 12 ÷ -5 = ¹²⁄₋₅
A Very Special Slope
y = x line has a slope of 1 and intersects at the origin.
It’s special because, in Quadrant I,
* For every point ABOVE this line the y-coordinate is always greater than the x-coordinate (y > x)
* For every point BELOW this line the x-coordinate is always greater than the y-coordinate (x > y)
Reflections Across Lines
Reflection about/across the y = x line : Simply swap the coordinates. (3, 2)→(2, 3)
Reflection about/across the y = -x line : Swap the coordinates and sign. (-3, 2)→(-2, 3)
Reflection about/across the y = k line (k is a constant) : Keep x coordinate, y-coordinate is (2k - y)
Reflection about/across the x = k line (k is a constant) : Keep y coordinate, x-coordinate is (2k - x)
Slope-Intercept Form
If you can convert a linear equation in the below form, it gives you a ton of information
y = mx + c
* m = slope
* c = y-intercept
* y = y-coordinates
* x = x-coordinates
Original equation: -5x + 3y = 6
-5x + 3y = 6→3y = 6 - 5x→ y = ⁵⁄₃x + 2
* Slope = ⁵⁄₃
* y-intercept = 2
Calculating next coordinate point from Slope-intercept form
Next coordinate point for
y = ⅔x + 1
y = 4x + 3
- Take the y-intercept from the form and convert it into coordinates: (0, y-coordinate)
- Add the slope to the y-coordinate→y-coordinate of next point
- Add 1 to the x-coordinate→x-coordinate of next point
y = 4x + 3→(0+1, 3+4)→(1,7)
- If the slope is a fraction, add the numerator to the y-coordinate and the denominator to the x-coordinate (0)
- If the slope is negative, subtract slope from y-coordinate
- y-intercept is 1, so coordinates are (0,1)
- Add 2 to the y-coordinate: 1 + 2 = 3 (y-coordinate of next point)
- Add 3 to the x-coordinate (0) : 0 + 3 = 3 (x-coordinate of next point)
→Coordinate of the next point is (3,3)
Calculate x-intercept from Slope-intercept form
What is the x-intercept of y = 3x + 1
Set y = 0 and calculate x.
- y = 3x + 1
- 0 = 3x + 1→1 = 3x→1÷3
- x-intercept = ⅓
Parallel Lines
What is a line parallel to 6x - 2y = 10?
Two lines that never touch each other.
In coordinate geometry, you know that two lines are parallel if they have the same slope and different y-intercepts.
6x - 2y = 10
* Convert to slope-intercept form: y = 3x - 5
* Slope = 3
* Every line with a slope of 3 with a different y-intercept is parallel: y = 3x + 7, y = 3x - 100
Perpendicular Lines
What is the slope of a line that is perpendicular to the line with the equation 4x -5y = 20
- Two lines that when they intersect, they form a 90° angle/Two lines whose slopes are negative reciprocals of each other.
- To find the reciprocal of a number, you simply “flip” it.
For example, the reciprocal of 3 = ⅓ - With the negative reciprocal, you not only have to “flip” it but change the sign as well.
For example, the negative reciprocal of −5 = ⅕
- Convert into slope-intercept form: y = ⅘x + 4
- Take the slope and flip it: ⅘→⁵⁄₄
- Make it the opposite sign: ⁵⁄₄→-⁵⁄₄
A line with a slop of -⁵⁄₄ is perpendicular
Using Systems of Equations to find the intersection point of two lines
Use systems of equations to find the intersection point of x + 4y = 11 and -2x + 3y = 0
Use systems of equations to find the intersection point of the two lines
1. Try to eliminate either x or y through systems of equations
2. Calculate the other by substituting the variable in one of the original equations
3. x = x-coordinate, y = y-coordinate of the intersection point
- Multiply x + 4y = 11 by 2 to eliminate x: 2 (x + 4y) =11→ 2x + 8y = 22
- Add (2x + 8y = 22) + (-2x + 3y = 0) = 11y = 22
- Calculate y: 22 ÷ 11 = 2→y = 2
- Substitute y = 2 in one of the original equations to get x: x + 4(2) = 11→x + 8 =11→x = 3
- Since x = 3 and y =2, the coordinates of the intersection point is (3,2)
Find the intersection point of two lines
Find the intersection point of 2x + y = 6 and x + 2y = 5
Another method other than System of Equations:
1. Arrange all the terms to one side to make each equation equal = 0
2. Set two equations equal to each other and single out the x to find equation for x.
2. Input the x-value and solve for y-coordinate
3. Input y-value to solve for x-coordinate
- 2x + y = 6→y = 2x + y -6 = 0
x + 2y = 5→x + 2y -5 = 0 - Set the equations equal to each other: 2x + y -6 = x + 2y -5
- Solve for x equation: x = y + 1
- Input x-value in one of the original equations to solve for y-coordinate: (y+1) + 2y = 5→y = ⁴⁄₃
- Input y-value to solve for x-coordinate: ⁴⁄₃ + 1 = 5→x = ⁷⁄₃
- Coordinates of intersection point = (⁷⁄₃, ⁴⁄₃)
Line Equations and their graphs
k is a positive constant
* x² + y² = k²→A Circle
* x + y = k→A Line
* y = kx²→A Parabola
* x²/a² - y²/b² = 1→Hyperbola
Quadratic Equations on the Plane
y = x²
Becomes a parabola
- If the x is neg, then the parabola will be upside down: y = -x²
- y = x² + 1: The parabola will have the curve point at 1 (1 is the y-intercept)
- x = y²: The parabola will face to the right
- x = -y²: The parabola will face to the left
- y = (x + 1)² : If it is in (), the curve point is opposite. Parabola will move to the left so the curve point is at -1
- Smaller value of x: Makes the parabola wider
- Larger value of x: Parabola is thinner
- Parabolas can have 0,1 or 2 x-intercepts
Graph Quadratic Equations by Factoring
y = x² + 8x + 15
y = x² + 8x + 15
1. Factor the equation: y = (x+3)(x+5)
2. Set y = 0 to calculate x-intercepts: x = -3 and -5 and plot on graph
3. To find the middle point/vertex, first we find the x-value by calculating the average of the x-intercepts: (-3 + -5) ÷ 2 = -4
4. Plug the average into the original equation to find the y-value: y = (-4)² + (8 x -4) + 15 = -1
5. So the coordinates of the vertex is (-4,-1)
6. Plot on graph and connect the dots
If the equation is not factorable, use “completing the square” to make it factorable and solve as above
Finding the Minimum Value/Vertex
y = - (x + 3) (x - 7)
y = (x + 5)² - 23
y = 2x² - 4x + 1
- Factor the equation:
If it has 1 x-intercept, that is the min value
If it has 2 x-intercepts, find the average of the 2 intercepts
y = -(x + 3) (x - 7)
1. Since there are two x-intercepts: -3, 7, we find the average to get the x-value: -3 + 7 ÷ 2 = 2
2. Then substitute x-value to find y-value: -(2 + 3)(2 - 7) = 25
→Vertex= (2,25) - Complete the square:
The min value is the constant (coordinates: opposite of x-value, constant)
y = (x + 5)² -23→(-5,-23)
* Vertex Form: y = ax² + bx + c
1. First, find the x-value with: -b÷2a
1. The substitute x-value in original equation to solve for y-value.
1. y = 2x² -4x + 1→-(-4)÷2(2) = 1
→2(1)² -4(1) + 1= -1
→Vertex = (1,-1)
Find the # of x-intercepts using Discriminant
We can use the discriminant to find the number of x-intercepts an equation has.
* If pos: 2 x-intercepts
* If 0: 1 x-intercept
* If neg: No x-intercepts
Graphing Circles
- 16 = (x−5)² + (y+2)²
- 3 = x² + 4x + y − 6y²
Circle Formula:
r² = (x - h)² + (y - k)²
* (h,k): Circle’s center
* r: Radius of the circle
* (x,y): Points on the circle
If the equation is r² = x² + y², it means that the center is at the origin (0,0)
If it is not in the above format, you might have to complete the square
- radius = 4, center = (5,−2).
- radius = 4, center = (-2,3)
Graphing Linear Absolute Value Equations
y = |x + 3|
- Split the linear absolute value equation into its two separate equations: pos and neg
- Find the x value that makes the “stuff” inside the absolute value lines equal to 0
- Plot BOTH equations from step 1, using the x-value in step 2 as the boundary point, or the “v”-point of the graph.
Don’t allow any neg y-values
- Split equations: y = x + 3, y = -x - 3
- Find boundary point: x = -3
- Plot the boundary point: (-3,0)
- For x-values≥-3: Plot y = x + 3
For x-values<-3: Plot y = -x - 3
Graphing Quadratic Absolute Value Equations
y= |x² − 4|
- In some cases, absolute value signs on a quadratic don’t change anything.
y = x² + 3
y = |x² + 3|
The two equations are equal to each other because it is impossible for y = x² + 3 to be negative. In this case, we can graph the quadratic as is - Many quadratic equations CAN spit out negative values.
1. Split the quadratic absolute value equation into its two separate equations and find the x-values.
2. Identify the two values as the “boundary points,” the points where the graph suddenly shifts directions.
3. Plot BOTH equations
Quadratic Absolute Value graphs look like a W when pos and an M when neg
- Split equation into pos and neg and calculate x-value (what makes y=0)
Value 1: x=2
Value 2: x=−2 - Identify the two x-values as the “boundary points,” the points where the graph suddenly shifts directions.
- For x values ≤−2 and ≥2, graph pos equation, making sure not to go beyond the boundary points.
- For x values between −2 and 2, graph neg equation
Rotating 90°
What is (3,2) rotated 90°?
When rotating a point 90°, the next point has to be on a line that is perpendicular to the first.
To find the new point, you switch the coordinates: (x,y)→((y,x) but you have to pay attention to what quadrant you’re in to determine whether the x-coordinate, y-coordinate, or both are positive or negative.
- If a point in Quad 1 is rotated 90° clockwise, it will be in Quad 4: (pos,neg)
- If a point in Quad 1 is rotated 90° anti-clockwise it will be in Quad 2: (neg,pos)
- 90° clockwise: (2,-3)
- 90° anti-clockwise: (-2,3)
