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# Gcse Maths- turning point watch

1. How would you find the turning point of a quadratic graph when you are only given the y intercept(0,-10) and the coordinate of one intersection on the x axis (5,0)? Any help would be appreciated
2. BTW just to add this was a question from the maths GCSE calculator paper 3 from 2017
3. (Original post by Dan_Jenkins)
How would you find the turning point of a quadratic graph when you are only given the y intercept(0,-10) and the coordinate of one intersection on the x axis (5,0)? Any help would be appreciated
Can you post the exact question or a picture? I looked in my copy of Edexcel Maths paper 3 2017 9-1 and I can't find it lol. I think I could solve it if it stated that a (the x squared co-efficient) = 1.

For the case that a = 1:

Edit: According to somebody in another thread, my solution is correct.
Spoiler:
Show

If 5 is a root of the quadratic (x-5) must be a factor:
(x-5)(x+k) = 0

-5 and k must multiply to give -10 as -10 is the y intercept.
-5k = -10
k = 2
(x-5)(x+2) = 0
x^2 -3x -10 = 0
By completing the square:
(x-1.5)^2 - 2.25 - 10 = 0
(x-1.5)^2 - 12.25 = 0

So x= 1.5, y = -12.25
Turning point = (1.5, -12.25)
4. (Original post by Loci Pi)
Can you post the exact question or a picture? I looked in my copy Edexcel Maths paper 3 2017 9-1 and I can't find it lol. I think I could solve it if it stated that a (the x squared co-efficient) = 1.

For the case that a = 1:

If 5 is a root of the quadratic (x-5) must be a factor:
(x-5)(x+k) = 0

-5 and k must multiply to give -10 as -10 is the y intercept.
-5k = -10
k = 2
(x-5)(x+2) = 0
x^2 -3x -10 = 0
By completing the square:
(x-1.5)^2 - 2.25 - 10 = 0
(x-1.5)^2 - 12.25 = 0

So x= 1.5, y = -12.25
Turning point = (1.5, -12.25)
Please do not post full solutions on the maths forum
5. (Original post by Y11_Maths)
Please do not post full solutions on the maths forum
Sorry, I will spoiler it now but it was only the solution for if a = 1, which the thread starter did not specify.
6. (Original post by Y11_Maths)
Please do not post full solutions on the maths forum
why not?
7. (Original post by misspessimism)
why not?
So the thread starter can attempt the question once they are given a few hints without the answer being spoiled.
8. (Original post by Loci Pi)
So the thread starter can attempt the question once they are given a few hints without the answer being spoiled.
Ahh. i thought you meant legal reasons
9. (Original post by misspessimism)
why not?
It is against the rules in the maths forum because this is not a “do my homework” place.
10. (Original post by Dan_Jenkins)
How would you find the turning point of a quadratic graph when you are only given the y intercept(0,-10) and the coordinate of one intersection on the x axis (5,0)? Any help would be appreciated
https://www.thestudentroom.co.uk/sho...4#post76280584

I've already given a solution for this problem for someone else...
11. (Original post by Y11_Maths)
It is against the rules in the maths forum because this is not a “do my homework” place.
Ah well thats my question out the window..
12. (Original post by Mehru1214)
https://www.thestudentroom.co.uk/sho...4#post76280584

I've already given a solution for this problem for someone else...
How are you finding my paper?
13. (Original post by Y11_Maths)
How are you finding my paper?
It looks good, however unfortunately I had to leave it, so I'm going to do it properly tomorrow! Nice though!
14. (Original post by Loci Pi)
Can you post the exact question or a picture? I looked in my copy of Edexcel Maths paper 3 2017 9-1 and I can't find it lol. I think I could solve it if it stated that a (the x squared co-efficient) = 1.

For the case that a = 1:

Edit: According to somebody in another thread, my solution is correct.
Spoiler:
Show

If 5 is a root of the quadratic (x-5) must be a factor:
(x-5)(x+k) = 0

-5 and k must multiply to give -10 as -10 is the y intercept.
-5k = -10
k = 2
(x-5)(x+2) = 0
x^2 -3x -10 = 0
By completing the square:
(x-1.5)^2 - 2.25 - 10 = 0
(x-1.5)^2 - 12.25 = 0

So x= 1.5, y = -12.25
Turning point = (1.5, -12.25)

AKA me XD
https://www.thestudentroom.co.uk/sho...4#post76280584
15. (Original post by Mehru1214)
It looks good, however unfortunately I had to leave it, so I'm going to do it properly tomorrow! Nice though!
Ok

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