Does anyone know where I can find practice questions involving parabolas?
Hi Kieran
I am assuming you are doing Tables Equations and Graphs Level 1 here is the link to last year’s paper. Please let me know if you need any help with it.
Mrs H
Level 1 Mathematics and Statistics (91028) 2022 (nzqa.govt.nz)
@Kieran, have a look at the following links:
You can also go through past papers, all of them include questions about quadratic functions.
Thanks so much. I’m having trouble with question b(i). Could you please explain how I can find q?
In one of the previous discussions for this topic we published some very helpful recourses about transformations of parabolas:
The Parabola.pdf (477.6 KB)
From this resource we can see that vertical transformation of the parabola from the origin (0,0) is defined by the number you add to x^2:
From your image you may notice that the vertex of the parabola is located at the point (0,9) so it was translated up by 9 units:
That means your equation will take the form:
So q = 9
Thanks, I did manage to figure it out. I just had to reread the question. There is another question on the test that I’m really stuck on which is b(iii). I’ve found a = 0.25 using the equation: y = a(x-h)^2 + c. However, I’m not sure if this is correct and I also don’t know what to do from there (assuming it is correct) to find the maximum height. Could you please explain it to me?
For question b(ii) to find the coefficient of x^2. The best way is to substitute the coordinates of known points into your equation:
Question b(iii) would be an excellence question, @Kieran. To find the equation of a new parabola you will need to use the symmetry property of the parabola:
You can state in your assessment that using the symmetry property of the parabola, if the graph crosses the x-axis at (6,0) and has a vertex at x = 0.5, then it must also cross the x-axis at (-5,0). Therefore, the equation of the parabola can be written using its x-intercepts:
Then, we can use the same method to find the coefficient a as we did for b(ii) - we will substitute the coordinates of a known point. Any point on the x-axis will make the equation equal to 0 and won’t help us, so we can use the y-intercept (0,9):
and your final equation will be:
To find the maximum height we need to substitute the x-coordinate 0.5 into our final equation:
m.
Thanks so much, that’s really helpful. Just a couple of questions with b(ii), firstly wouldn’t the equation be y = - ax^2 + 9 because the parabola is a n shape and not a u shape? Secondly, would it be better to write a as the fraction or the decimal? Thanks again for your help.
Good questions, @Kieran. We can write our equation in the general form y = ax^2 + 9 (as I did) and then we will get a equal to a negative number. If you correctly assume that a will be negative and write your general form as y = -ax^2 + 9, your answer for a will be positive. As long as the final equation is correct, it doesn’t matter which way you came to this solution.
For the second question - it also doesn’t matter in which form you write your answer - fraction or decimal. If you choose to use decimal, round your answer to at least 3 or 4 d.p.
Alright, thanks so much for your help.
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