Chapter P

P.1

 Order Pairs (x,y)

X represents the distance of the point from the Y axis

Y represents the distance of the point from the X axis

Quadrants 

Translating Points

Original Points Transformed Point 

(x,y)     (-x,y)    reflection of the original point in the Y-axis

(x,y) (x,-y)    reflection of the original point in the X-axis

(x,y) (-x,-y)   reflection of the original point through the   origin

Distance Formula 

Given the two points (x₁, y₁) and (x₂, y₂), the distance between these points is given by the formula:

• 

Midpoint Formula

If the coordinates of A and B are ( x₁, y₁) and ( x₂, y₂) respectively, then the midpoint, M, of AB is given by the following formula (Midpoint Formula). 

Standard Form of the Equation of a Circle 

The point (h,k) is the center of the circle, and positive number r is the radius of the circle. The standard form of the equation of a circle whose center is the origin is 

 

P.2

Determining Solution Points 

The point (2,13) lies on the graph of y = 10x-7 because it is a solution point of the equation 

y = 10x - 7

13 = 10(2) - 7

13 =13

How to Sketch the Graph of an Equation by Point Plotting 

1.If possible, rewrite the equation so that one of the variables is isolated on one side of 

the equation.

1. Make a table of several solution points.
2. Plot these points in the coordinate plane.
3. Connect the points with a smooth curve. 

Using a Graphing Utility to Graph an Equation 

1. Rewrite the equation so that y is isolated on the left side.
2. Enter the equation into a graphing utility.
3. Determine a viewing window that shows all important features of the graph.
4. Graph the equation. 

P.3

Slope

The slope m of the non vertical line 

Point - Slope Form

y – y₁ = m(x – x₁)

Slope - Intercept Form

y = mx + b

Parallel lines are only parallel if the slopes are equal.

Perpendicular lines are only perpendicular only if the slopes are negative reciprocals. 

P.4

Solving an Equation

2x² - 5x - 12 = 0.

(2x + 3)(x - 4) = 0.

2x + 3 = 0 or x - 4 = 0.

x = -3/2, or x = 4.

Finding X and Y intercept 

• y = 0 for the x-intercept(s), so:

• 25x² + 4y² = 9
  25x² + 4(0)² = 9
  25x² + 0 = 9
  x² =  ⁹/₂₅
  x = ± ( ³/₅ )

• x = 0 for the y-intercept(s), so:

• 25x² + 4y² = 9
  25(0)² + 4y² = 9
  0 + 4y² = 9
  y² =  ⁹/₄
  y = ± ( ³/₂ )

Points of intersection

y = 3x + 2, x = 3 gives
y  =  3(-3) + 2
y  =  9 + 2
y  =  7

y  =  2x - 1, x = -3 gives
y  =  2(3) – 1
y  =  6 – 1
y  =  7

Thus, the point (3, 7) is the point of intersection of the two lines. 

Solving quadratics 

Factoring 

x2 + 5x + 6 = x2 + 3x + 2x + 6       = (x2 + 3x) + (2x + 6)       = x(x + 3) + 2(x + 3)       = (x + 3)(x + 2)

Completing the Square 

(x – 4)² = 5
x – 4 = ± sqrt(5)
x = 4 ± sqrt(5)
x = 4 – sqrt(5)  and  x = 4 + sqrt(5)

Quadratic Formula 

Solving Equation Absolute Value 

• Solve | x + 2 | = 7
• To clear the absolute-value bars, I must split the equation into its two possible two cases, one case for each sign:

• (x + 2) = 7     or     –(x + 2) = 7
  x + 2 = 7       or     –x – 2 = 7
  x = 5             or     –9 = x

P.5

Solving  Inequalities

7 - 2x < 3

-2x < -4

x > 2

Solving Double Inequalities 

Solving Absolute Value 

| 2x + 3 | < 6
–6 < 2x + 3 < 6     
–6 – 3 < 2x + 3 – 3 < 6 – 3
–9 < 2x < 3
^–9/₂ < x < ³/₂