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INSTRUCTORS Raffi Hovasapian John Zhu
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For more information, please see full course syllabus of Calculus AB
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Lecture Comments (14)

1 answer

Last reply by: Nolan Zhang
Tue Apr 7, 2015 8:14 PM

Post by Nolan Zhang on April 7, 2015

Good evening, so by what is the definition of 'reflect something over X/Y axis as in example 3'. It seems to me that you simply draw out what G(x) looks like using the given info. So does 'reflect' simply means 'draw something out?'

1 answer

Last reply by: John Zhu
Thu Jan 15, 2015 11:12 AM

Post by abendra naidoo on December 13, 2014

should not g(x) be g(x)=-2f(x^2+2)+1 instead of just x as you have written?

1 answer

Last reply by: John Zhu
Thu Jan 15, 2015 11:00 AM

Post by Rebecca Dai on November 4, 2014

Hi, if I want to do Calculus BC, should I watch the AB first and then watch the BC? Is that enough if I just finished pre-calc? Thanks!

0 answers

Post by Nada Al-Hellai on October 30, 2013

Hello ;3

Why in example5 1/2|x+2| the +2 was a H.shift.... shouldn't it be vertical? And if it's a H.shift.. why then example6 X^3+1 the +1 is a V.shift?
What is the difference and how do you determine that?

3 answers

Last reply by: David Bush
Thu Aug 29, 2013 7:57 PM

Post by Tanveer Sehgal on November 20, 2012

Hey,

if g(x)=f(|x|).... Then the graph created is a parabola to the right. In the case if we do the vertical line test, wont the graph no longer be a function? So wont g(x) not be a function?

0 answers

Post by Kusuma Cherukuri on June 3, 2012

How do you know whether to graph the shift first or the stretch first. Is there a set order to graphing the manipulations or do you just go in order of how it is written?

1 answer

Last reply by: Derrick Lu
Thu Jun 28, 2012 5:20 PM

Post by Kusuma Cherukuri on June 3, 2012

Why can't y = 0 just because x can't equal 0? Couldn't y = 0 if x = -1/2? This would be the ordered pair for y = 0: (-1/2,0). Right?

Related Articles:

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.

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