Aisha W.
asked • 08/29/23f(x)=|x+3|; horizontal stretch by a factor of 4
More
3 Answers By Expert Tutors
Yefim S. answered • 08/29/23
Tutor
5
(20)
Math Tutor with Experience
G(x) = g(x/4) = Ix/4 + 3I
Dayv O.
G(x)=0 at x=-12, g(x)=0 at x=-3. Yes g(x) stretched by 4 with G(x), but also G(x) is g(x) translated -9 on x-axis which was not part of problem. Treating x+3 as x' and then G(x')=g(x'/4) works.
Report
08/29/23
Dayv O.
say problem was stretch g(x)=sin(x-pi/4) by four horizontally. G(x)=sin((1/4)(x-pi/4)) NOT G(x)= sin((x/4)-pi/4).,,,,,True?
Report
08/29/23
Roger R.
tutor
@Dayv: No, that's not true. The point (π/2,1/√2) on y = g(x) will become (2π,1/√2) after the horizontal scaling by 4 and satisfies y = sin(x/4−π/4), but not y = sin([x−π/4]/4). The point (0,−1/√2) on y = g(x) becomes, well, (0,−1/√2) and does, obviously, not satisfy y = sin([x−π/4]/4).
Report
08/29/23
Dayv O.
let's graph h(x)=sin((1/4)*(x-pi4)) and k(x)=sin(x-pi/4), both have x axis intercept at x=pi/4. If x=pi/2 in h(x), h(x)=sin(pi/16). If x=pi/2 in k(x), k(x)=sin(pi/4). Now let's graph m(x)=sin((x/4)-pi/4). The graph of m(x) is translated 3pi/4 positive on x-axis. The points (pi/2,1/sqrt(2)) and (2pi,1/sqrt(2)) are measured from x=0, not from x=pi/4, and parent function of sin(x) was translated with k(x)=sin(x-pi/4)
Report
08/29/23
Dayv O.
@Roger: from above comment: h(x)=1 when x= 9pi/4,,which is 2pi from x=pi/4. k(x)=1 when x=3pi/2,,which is pi/2 from x=pi/4. function stretched by four horizontally.
Report
08/29/23
Dayv O.
recalling electric engineering principles, h(x) is only a frequency change to k(x) and meanwhile m(x) is both a frequency change and a phase shift of k(x).
Report
08/30/23
Roger R.
tutor
“[H]orizontal stretch by a factor of 4” means stretching from x = 0. The y-axis stays put, not the line x = π/4. It's the map (x,y) ↦ (4x,y).
Report
08/30/23
Roger R.
tutor
The whole point of the exercise is that an affine transformation (x',y') = T(x,y), applied to the curve k: f(x,y) = 0, works (algebraically) directly (!!) at the variables (x,y), k': f(S(x,y)) = 0 where S = T⁻¹, the inverse transformation. The inverse! That's why y = f(x−3) is y = f(x) shifted to the right while algebraically it seems shifted to the left. [A point (x',y') on the new curve is moved to the left (x'−3,y') to lie on the old curve with the equation y = f(x), so plug in the coordinates: y' = f(x'−3) is the relation between the coordinates (x',y') on the new curve!]
Report
08/30/23
Dayv O.
great, when stretching ellipse [(x+3)^2/9]+y^2/4-1=0 by factor of 4 horizontally, we gleefully say the new ellipse is [((x/4)+3)^2]/9+y^2/4-1=0 is new curve. yikes, the ellipse moved and is elongated when plotted directly(!!). But the plot of [((x+3)/4)^2]/9]+(y^2/4)-1=0 is exactly only horizontal stretching. I thought problem was to stretch out y=|x+3| horizontally by factor of 4, stretching the plot and not moving the plot.
Report
08/30/23
Dayv O.
The graph of y = A*f[B(x + h)]+k,,,for constant B: B: The function stretches or compresses horizontally by a factor of 1/|B|. If B is negative, the function also reflects across the y-axis. Please see pdf caraleemath.com/uploads/2/8/4/0/28409927/section_1.5_order_of_transformations_and_summary.pdf We teach this in pre-calculus. Yes?
Report
08/30/23
Roger R.
tutor
I don't quite understand how you relate an English phrase like “The function stretches or compresses horizontally by a factor of 1/|B|” to the mathematical/geometric operation of scaling. What does B = 1/2 exactly do in the equation y = √{1/2⋅(x +5)}? Compared to what graph y = f(x)?
Report
08/30/23
Dayv O.
I really don't understand, this is exactly what is taught in pre-calculus textbooks. Have some parent function,,,y=f(x), then clearly every time in text books they explain y = A*f[B(x + h)]+k, not y = A*f[Bx + h]+k. because then x=-h/B not x=h would be translation on x axis. For this problem parent function is absolute value. f(x)=|x|. g(x)=|1*(x+3)| moves the function. h(x)=|(1/4)*(x+3)| stretches the moved function. Is this hard?
Report
08/30/23
Dayv O.
y=sqrt((1/2)*(x+5)) stretches horizontally by factor 2 the function g(x)=sqrt(x+5).
Report
08/30/23
Roger R.
tutor
Either your answer is plain false, or we two have a completely different understanding of "stretches horizontally by factor 2 the [curve] y=sqrt(x+5)".
Report
08/30/23
Roger R.
tutor
Maybe, you can explain how you "stretch horizontally by factor 2".
Report
08/30/23
Dayv O.
I graph f(x)=sqrt(x+5), it is 5 when x+5=25, I graph h(x)=sqrt((1/2)(x+5)), it is 5 when x+5=50. I guess there is a different higher math world where when applying a constant to domain to scale, it is no problem the graph has a different x-axis translation based on the scaling constant. affine math and inverses. In the practical world the scaling constant is applied to the variable contained within parenthesis so x-axis translation is maintained.
Report
08/30/23
Roger R.
tutor
You don't stretch anything. You plot two graphs by solving equations or plugging in numbers and then call one "stretched by factor 2". It would help if you drew the new curve using only a ruler to measure segments. "Stretching horizontally by 2" means certain segments/inches were doubled. Which segments? Answer: The distance from (x,f(x)) to the y-axis. The image of P(3,f(3)) on curve k is P'(6,f(3)) on curve k'. Q(−2,f(−2)) is mapped onto Q'(−4,f(−2)), etc. That's the conventional geometric procedure for "stretching horizontally by factor 2".
Report
08/30/23
Roger R.
tutor
If you call k': y = √{1/2⋅(x+5)} "stretched horizontally by factor 2" from the graph k: y = √{x+5}, you're not following the conventional stretching procedure. The point (−5,0) on k should map onto (−10,0). The latter point is not even on k'! You are stretching the distances to the line x = −5 and map, e.g., (20,5) on k [distance = 20+5] to (45,5) on k' [distance = 45+5]. That's okay, but don't expect others to understand you. You're singling out a particular point on the graph as the reference for the stretching procedure. That's not the conventional way to do it.
Report
08/30/23
Roger R.
tutor
For an unsuspecting reader, your statement "y = √{1/2⋅(x+5)} is stretched horizontally by factor 2 from ..." is just false.
Report
08/30/23
Dayv O.
and your implied argument that changing shape of ellipse (x+3)^2/9+y^2/4=1 so it is elongated by factor of 4 horizontally is equation ((x/4)+3)^2/9+y^2/4 is blatantly wrong. Now the center of the ellipse is moved from (-3.0) to (-12,0,. just so an arbitrary relation is maintained (changing shape is measured from point (0,0)). Is the center of ellipse ---- singling out a particular point ---?
Report
08/30/23
Roger R.
tutor
Again, Dayv, this is a language issue. The instructions on what you want to do must be as unambiguous as possible. If you say stretch the ellipse horizontally by factor 4, I understand it to move the center from (−3,0) to (−12,0). If you say: Find the equation of an ellipse with the same center (−3,0) but the horizontal semi-axis 4⋅3 (instead of 3), that's a completely different task. I would stretch w.r.t. x = −3.
Report
08/30/23
Dayv O.
There are I would acknowledge probably plenty of unsuspecting readers.
Report
08/30/23
Dayv O. answered • 08/29/23
Tutor
5
(55)
Attentive Reliable Knowledgeable Math Tutor
Yefim's answer is correct in that a unit increase of x/4 requires x=4. But because
in f1(x)=|x+3| the function is translated left by 3 must apply (1/4) to x+3
function stretched by factor of four is
f2(x)=(1/4)|x+3|=|(x/4)+3/4|
f1(-3)=0 and f2(-3)=0 so f2 only stretches by factor of 4.
Problem only wanted stretching, not moving x-axis intercept
Problem is not asking function to be translated.
BILAL S. answered • 08/29/23
Tutor
5.0
(71)
Helping Students Master AP Physics & Math with Confidence
Hello Aisha;
If we want to stretch it horizontally;
g(x) = |(x+3)/2|
I hope the Desmos link works
https://www.desmos.com/calculator/oyn3mgfb0v
Dayv O.
why 1/2 and not 1/4?
Report
08/29/23
BILAL S.
tutor
It 's my typo Dayv, for sure it is 1/2. Apologies
Report
08/29/23
Dayv O.
my question is the answer should be (1/4)|x+3| not (1/2)|x+3|.Agree?
Report
08/29/23
BILAL S.
tutor
Yes, (1/4)|x+3|
Report
08/29/23
BILAL S.
tutor
Thanks for following up.
Report
08/29/23
Dayv O.
no square (=no problem)
Report
08/30/23
Still looking for help? Get the right answer, fast.
Ask a question for free
Get a free answer to a quick problem.
Most questions answered within 4 hours.
OR
Find an Online Tutor Now
Choose an expert and meet online. No packages or subscriptions, pay only for the time you need.
BILAL S.
08/29/23