INSTRUCTORS Raffi Hovasapian John Zhu

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For more information, please see full course syllabus of Calculus AB

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### Graphing

• Avoid confusion: focus on 1 modification at a time

### Graphing

The following equation is that of an circle with its center at [0,0]. Rewrite the equation so that it's the same circle, but centered at [2, −1]
x2 + y2 = 1
• We want to shift x to the right 2 units, and y to the left 1 unit. That means when we plug in [2,−1] we get the same results as when we plugged [0,0] into the original equation
• 02 + 02 = 0
(x − x1)2 + (y − y1)2 = 0
(2 − 2)2 + (−1 + 1)2 = 0
(x − 2)2 + (y + 1)2 = 1
The following is an equation of a parabola y = x2 + 4x + 5 Alter the equation so that the parabola opens in the opposite direction, while keeping the same vertex
• y = x2 + 4x + 5
y = x2 + 4x + 4 + 1
y = (x + 2)2 + 1
f(x) = −(x + 2)2 + 1
Graph the following function f(x) = |x|
• This is equivalent to the piecewise function
• f(x) = {
 x : x ≥ 0
 −x : x < 0
Graph the following equation y = lnx
• Let's remember what lnx means. The equation can be rewritten as x = ey. There's no real number you can raise e to to get 0 or less than 0. x must be greater than zero for lnx to remain real. Domain (0, ∞). x approaches 0 as y approaches negative infinity and it is an increasing function. Range (−∞, ∞). It can also be seen that when y = 0, x = 1
0.0  8
Graph the following function f(x) = |x3|
• This graph will be the same as x3 when x ≥ 0, for x < 0, it will be the same, but flipped about the x axis
Graph the following function f(x) = 2√{|x+1|} − 3
• √u is an increasing function that will have its minimum when u is zero
• Plugging in some values:
[−1, −3]
[0, −1] [−2, −1]
[3, 1] [−5, 1]
[8, 3] [−10, 3]
Graph the following function g(x) = |x|2
• Here, the absolute value sign is redundant. Let's take a look at f(x) = x2
• f(x) = x2
f(−x) = (−x)2
= (−1)2 x2
= x2 = f(x)
• f(x) = g(x) for all real numbers
Find the equation of the following graph
• Shape of a parabola, so it has to be in the form of y = x2. Vertex is at [−3, −1], y increases by 1 when going over 1 x unit from the vertex
y = (x + 3)2 + 1
Flip the following equation about the x axis y = (x + 1)2
• This means that whenever y would have been positive, it will be negative and vice versa
y = −(x + 1)2
Stretch the following graph vertically y = (x + 1)2 + 1, keeping the orientation and vertex the same
• All we have to do is multiply the squared amount by any number greater than 1. Let's try 3
y = 3(x+1)2 + 1

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

### Graphing

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

• Intro 0:00
• Manipulating 0:10
• A in the Equation
• B in the Equation
• C & D in the Equation
• Negative values
• Example 1
• Example 2 1:51
• Example 3: Absolute Value Functions 3:43
• Example 4
• Example 5 6:17
• Example 6 8:02
• Example 7 9:10
• Example 8 11:02
• Example 9 11:47