i need some help with questions as my teacher over-complicates everything, feel free to add your own questions so we can help each other!!
methods questions help URGENT
give a sequence of transformations which transforms the graph of y=x4 to the graph of y=-4(x+1)4+2.
i don't understand what the -4 transformation is, thank you in advance
the -4 represents 2 things, one is the factor of dilation and the other is a reflection on the x-axis. Since its negative it will flip, and it dilated by a factor of 4
thank you so much!!
can someone please explain synthetic division?
honestly, I never had to do synthetic division, if you get really good at long division and quick at it it wont make a difference.
c25 Personally, synthetic division is the method I use and it generally is quicker then long division. In Year 11 (I'm doing 3/4 methods now), we had to learn both synthetic and long division and we were then able to choose which method we preferred. I do agree with lil N here in saying that if you're quick at long division it wont make a difference, however in my experience, synthetic division is quicker and involves less working. That being said, if you like to map out your steps so you don't lose track of your working, long division may be better suited to you.
These are some good videos on synthetic and long division to assist in learning the methods:
Synthetic: https://www.youtube.com/watch?v=FxHWoUOq2iQ
Long division: https://www.youtube.com/watch?v=_FSXJmESFmQ
Both of these videos are from the Organic Chemistry Tutor, I'd highly recommend watching his videos for concepts in Methods that you don't understand as he explains them in a simplified way.
to add on to what novahkiin said, if you do specialist aswell long division will be better in the long run when you do long division with complex numbers, but its up to you.
17 days later
Hi! Could someone please help w this question?
Two taxi services use the following different systems for charging for a journey:
Gold Taxi Initial charge of $10, plus a charge of 50 cents for each 200 m travelled
Purple Taxi Flat fee of $25 for travelling up to 20 km, plus a charge of $1 for each
kilometre travelled beyond 20 km
a Let G(d) be the cost (in dollars) of a journey of d km in a Gold Taxi. Show that
G(d) = 2.5d + 10 for d ≥ 0
b Let P(d) be the cost (in dollars) of a journey of d km in a Purple Taxi. Show that
P(d) =
25 for 0 ≤ d ≤ 20
d + 5 for d > 20
c On the same coordinate axes, sketch graphs to represent G(d) and P(d).
d Find the cost of a journey of:
i 7 km in a Gold Taxi ii 12 km in a Purple Taxi
e Art wants to travel a distance of 15 km. Which taxi service will be cheaper?
f Find the distances for which a Purple Taxi is the cheaper option
- Edited
girlmeetsvce
This is what each what each part is asking and how you should go about answering each:
a) Using the information given for the Gold taxi, formulate a function that corresponds to the cost (dollars) in terms of d (km).
b) Using the information given for the Purple taxi, formulate a function (G(d)) that corresponds to the cost (dollars) in terms of d km. Note that unlike the Gold taxi, the Purple taxi fee has two conditions and therefore the function P(d) will be a hybrid/piecewise function as you must define both conditions.
c) Graph both the functions, pretty self explanatory, with d km on the x-axis and cost on the y-axis. I'd recommend consulting your CAS for this (usually these types of questions are tech enabled).
d)
i) you need to set d=7 as an input for the function G(d). G(d) means 'the cost of the Gold Taxi (dollars) with respect to distance (km)', so given km you can find the cost.
ii) similar to part i), you need to use d=12 as an input for the function P(d). However, since there are two conditions, you need to ask: into which domain will d=12 fit into? In this case, d=12 fits into the domain of P(d)=25, and thus you'll use that function to find your answer.
e) There are two ways you can answer this question: inspecting the graph drawn in part c) or setting d=15 as an input for both function (remember for function P(d) you must assess which domain this input fits into).
f) Go back to the graph drawn in part c) and assess: where is the cost ($) for the function P(d) lower than that of G(d). From what I can see by graphing this on my CAS is that the cost of the purple taxi is lower than that of the Gold taxi after the intersection point of the two graphs. Therefore, we can state the distances for which the Purple taxi is the cheaper option with a domain: d∈ (6,∞) or for d>6, where the value of d at the intersection point is the first value in the domain. This value however, is given a not inclusive sign as it is the point where the cost of the Gold and Purple taxi are the same, not where the purple taxi is cheaper.