# How do I find the values of a and b from this sketch of the quadratic function?

j(x) = ax^2 + bx + c
https://ibb.co/vDpw6yh
I found c to be -1, and the y value of the turning point to be 1.5. But I'm not sure how to find the x value and figure out how to work out a and b?
(edited 1 year ago)
Original post by Bleepbloopbop
j(x) = ax^2 + bx + c
https://ibb.co/vDpw6yh
I found c to be -1, and the y value of the turning point to be 1.5. But I'm not sure how to find the x value and figure out how to work out an and b?

Ok so roots for j(x) are -1 and 4 right.
Therefore j(x)=(x+1)(x-4). This works because if you made j(x) =0, you would get roots -1 and 4 as required.
However when you expand the brackets, you get the y-intercept, c, = -4 which is not the case because the y-intercept is -1.
If you think about it, when we stretch j(x) in the y direction (i.e. up or down), roots will remain the same.
Therefore let q be a transformation of j(x) and therefore j(x)=q(x+1)(x-4).
We know when x=0, j(x) = -1 (the y intercept) and therefore sub in x=0 into j(x) and make it equal to -1: -1=q(0+1)(0-4) and rearrange to find q.
q comes out as 1/4 which makes sense because if you stretched j(x) by scale factor 1/4 in the y direction, the roots will remain same but you will get a y-intercept of -1.
Now expand to get an and b .
Original post by Bleepbloopbop
j(x) = ax^2 + bx + c
https://ibb.co/vDpw6yh
I found c to be -1, and the y value of the turning point to be 1.5. But I'm not sure how to find the x value and figure out how to work out a and b?

Unless I'm missing something, you don't need the turning point - you're given 3 points on the graph, so ...
Original post by Bleepbloopbop
j(x) = ax^2 + bx + c
https://ibb.co/vDpw6yh
I found c to be -1, and the y value of the turning point to be 1.5. But I'm not sure how to find the x value and figure out how to work out a and b?

Besides the y-intercept, (0,-1) which you identified, you can use the remaining two points given on the graph to form two simultaneous equations by subbing the x and y values into y = ax^2 + bx -1 and solve for a and b.
Original post by davros
Unless I'm missing something, you don't need the turning point - you're given 3 points on the graph, so ...

Even given the factorisation...
(edited 1 year ago)
Original post by Sahij
Ok so roots for j(x) are -1 and 4 right.
Therefore j(x)=(x+1)(x-4). This works because if you made j(x) =0, you would get roots -1 and 4 as required.
However when you expand the brackets, you get the y-intercept, c, = -4 which is not the case because the y-intercept is -1.
If you think about it, when we stretch j(x) in the y direction (i.e. up or down), roots will remain the same.
Therefore let q be a transformation of j(x) and therefore j(x)=q(x+1)(x-4).
We know when x=0, j(x) = -1 (the y intercept) and therefore sub in x=0 into j(x) and make it equal to -1: -1=q(0+1)(0-4) and rearrange to find q.
q comes out as 1/4 which makes sense because if you stretched j(x) by scale factor 1/4 in the y direction, the roots will remain same but you will get a y-intercept of -1.
Now expand to get an and b .

Thank you, I wouldn't have thought of this but it makes sense
Original post by randomperson321
Besides the y-intercept, (0,-1) which you identified, you can use the remaining two points given on the graph to form two simultaneous equations by subbing the x and y values into y = ax^2 + bx -1 and solve for a and b.

I'm sorry I'm still sort of confused if I do that I get -a^2 - b - 1 = 0 and 4a^2 + 4b - 1 = 0, wouldn't the a's and b's just cancel each other out if I tried to solve for 1?
Original post by Bleepbloopbop
I'm sorry I'm still sort of confused if I do that I get -a^2 - b - 1 = 0 and 4a^2 + 4b - 1 = 0, wouldn't the a's and b's just cancel each other out if I tried to solve for 1?

You misread the equation, a isn't squared here
If you sub it in correctly, you get this:

Spoiler

Original post by randomperson321
You misread the equation, a isn't squared here
If you sub it in correctly, you get this:

Spoiler

Mb thank you very much for the help