What questions do you have about this external?

Hi all

If there are any areas you need help on for this standard please ask. We are happy to help explain answers to past papers or can answer other questions you are stuck on.

This is one of my favourite standards :slight_smile:

1 Like

Hi there,

Do you have any useful info on finding equations of exponential graphs? This standard is a part of my end of year exams and it’s been a while since I last revisited it. More than anything, what happens when y doesn’t equal to y=2^x, for example, y=3^x? What changes here?

Hi @saintella

This is a great question as these often come up in the exams! A lot of the time you will be required to draw these or translate these.

When it is in the form y=2^x the values in the graph will be doubling, whereas if it is y=3^x the graph will be tripling.

The graph can move in a similar way to the other graphs in this standard. For example y=2^(x+a)+b
The “a” part would be a left/right shift and the “b” part moves the graph up and down.

I can put together some examples for you and will upload this later. There is also some useful links here

Thanks for the question

1 Like

Here is the link to the notes I have made for my students:
Exponential summary
Parabola summary

1 Like

Thanks for this explanation, that was super helpful! I’ll check out the examples you uploaded too. I really appreciate it :slight_smile:

Another question I might add is what happens when an exponential function looks like this?:

Y= 18 x 1.6 ^ x-1
What do the two base values indicate?
The context is:
Bob is monitoring a sunflower plant’s growth for a biology experiment. For the first 6 weeks, the height (Y) in cm can be modelled by the function Y=18 x 1.6 ^x-1, where x is the number of weeks from the start.

This is a practice Q I have come across in an exercise book and I’m not exactly sure of where to begin with it.

Interesting question! Exponential function is very popular in Biology, Physics, Finance, etc and it is really useful to understand it deeply.
Let’s look at all these numbers and see what each of them means practically.

  1. The base of an exponential function (1.6 in our problem) tells us by how many times the function increases each time unit. In our case, the height of Bob’s plant is multiplied by 1.6 each week. Unlike linear growth when the increase is constant (e.g. 10cm each week) exponential function grows faster the bigger it is.
    If the plant started from 1 cm then in one week’s time it will be 1.6 cm, in two - 1.6 * 1.6 = 2.56 cm, in three - 1.6 * 1.6 * 1.6 = 4.096 cm and so on.

  2. We have our function multiplied by 18:
    Compare these two functions. The growth is the same, the functions are basically parallel, they are multiplied by 1.6 each week. However, the red function (y = 18 * 1.6^x) crosses the y-axis at 18 (when x = 0 y = 18 * 1.6^0 = 18) while the purple function - at 1 (x=0, y = 1.6^0 = 1). So the number you multiply the exponential function by (18) is the initial height of Bob’s plant, before he started observation, at week 0.

  3. There is one more number we didn’t look at and that is -1 in the power. Look at these functions in comparison:

    When we have x-1 instead of x in the power, it means Bob’s plant was 18 cm tall one week after observation started (x=1, y=18*1.6^(1-1)=18), not at week 0.
    Another way of looking at this x-1 is to apply Algebraic procedures:
    The initial height of Bob’s plant was 18/1.6=11.25 cm and it reached 18cm only in one week time after observation.

I hope that helps. Please, ask if you need clarification!

1 Like

Thank you! The algebraic procedures are a bit beyond what I have learned so far however should be standard when I come across this kind of problem. Thank you for clarifying which each part of the equation was relevant to and clarifying with diagrams! It was super helpful seeing it all work visually.

Look at these links about indices and how to deal with them:

I have another question regarding exponential functions- very similar to this one.
Here is the problem:
A strain of bacteria doubles every 3 hours. At the start of the experiment there are 50 bacteria present in a culture.
The rate of growth over a 12 hourly period can be given by the formula N=50x2^t/3 =, where N is the number of bacteria and t is the time in hours.

I was asked to draw a graph, so used your explanation of the last question to help me. I assumed that because the function was multiplied by 50, the y-intercept of this function would be y=50. It doubles every 3 hours, so 3 hours after x=0, x=3 and y=100 because 3 hours before that the culture had only 50 bacteria and double of 50 is 100. I wasn’t entirely sure of how this equation worked as now the power is x/3, rather than x-1. What does this mean?

Would it be safe to assume that the rest of the equation means:
N=50 x 2 ^ t/3
50 is the initial number of bacteria, y-int
2 is the number in which the function is multiplied by every few hours in this context
to the power of t/3 and t is the time in hours divided by 3 - this is the part where I’m confused and wouldn’t know how to explain or apply to a graph, or work out on a graphics.

You are doing so well progressing in this topic! It will really help you next years when/if you do Calculus.

You are absolutely correct. 50 would be your y-intercept, at t=0 (beginning of the experiment) you have 50 bacterial cells. In three hours it will be 100 cells (N = 50 * 2^(3/3) = 50 * 2 =100). In 6 hrs from the beginning of the experiment you will have 200 cells (N = 50 * 2^(6/3) = 50 * 2^2) and so on.

The division by 3 in your power tells you that the bacterial culture doubles every 3hrs (not every hour). Look at this graphs in comparison:

The blue graph has x in the power so you can see that the bacterial culture doubles every hour: 50 cells at the beginning of the experiment, 100 in one hour, 200 in two hours and so on.

Red line has x/3 in the power. It means that the graph grows three times slower. It reaches 100 cells only in three hours, 200 - in six hours and so on.

I hope that helps :slight_smile:

Have you used before? It’s free and very easy to use. You can see what your equation looks like graphically and that may help you to understand it better. If you haven’t yet try it!

1 Like

Thank you again so much! I see that now, it’s interesting to see the comparisons of the two questions when ^x-1 and ^x/3 and how that changes the shape of the graph in context. Thank you for taking the time to write this all out for me! I’ll definitely check out Desmos and write a few notes from those links you sent me regarding the algebraic expression. :grin:

1 Like

Hi there, I’m back with another question!
I have come across a question relating to parabolas. I was asked to find the equation of a parabola, and because the parabola does not have a clear turning point (the co-ordinate said -1.5, -6.25) I used the intercept method and got the equation of y=(x-1)(x+4) which I believe is correct.
Here is the parabola:

The next part of the question asks me to translate this parabola 2 units to the right, and 3 units up and to provide an equation.
My question is, how am I able to translate a parabola using the intercept method equation model if I don’t yet know what the new intercepts are?

Thanks :slightly_smiling_face:

Good questions @saintella. The short answer - you can’t. Well, you can but you will need to re-arrange your equation to bring it to the form of y = a(x - h)^2 + k.

Your equation based on x - intercepts is correct but you can’t use it in this form to find equation for your new parabola. If you write this equation in the complete square form it will be

y = (x+1.5)^2 - 6.25.

That is the same equation you had just in different form. There is a way to convert equations from one form to another but I believe it is outside of this course.

To move your parabola 2 units to the right and 3 units up you have to do the following:


1 Like

Thanks for that! I’ll just do my best to keep that idea in mind if I come across anything like this. The activities I was using to practice are quite difficult and don’t reflect the paper entirely so I’m hoping that this sort of question won’t appear on the exam.

I did however come across a question I have seen frequently across all exams for TEG1.3 and I struggle with this one the most.

Ignore my working however you will see my attempts and mistakes already. My exam is tomorrow morning, so this really is quite last minute but any feedback at all for this type of question would be amazing. I really struggle with the context of the equation. I like to know what each part of the equation means and I’m struggling to apply my prior knowledge to this one.


You are doing so well @saintella! You identified all key coordinates correctly and answered the first question absolutely right. Because x is the distance from point A we conclude that at A x=0. If we substitute 0 into our equation we will get y=0/3 * (0 - p) + 4 = 4 therefore y (the height of the rope above the ground) is 4m.

Secondly, we need to find the value of p. Your approach is absolutely correct. You substituted the coordinates of the lowest point (3,1) into your equation to find p. However something went wrong with the calculation. If you solve the equation p = 6:

So the equation should be y = x/3 * (x - 6) + 4. As you can see from the diagram the height of the rope is equal to 4m in two cases: when x = 0 (point A) and when x = 6 (point B). So you could probably even guess from the diagram that p=6 as that would make (x - p) = 0 so y(the height) = 4.

The graph should look like this:

Good luck for your exams tomorrow!