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1. The equation of a graph is of the form y=x^2 +bx+c.
The graph crosses the x- axis at -4 and 5
Sketch the graph and find the values of b and c

I know the answer is c = -20 and b=-1 but I'm not entirely sure how to work it out. Can anyone help
2. Y=(x 4)(x-5) they are the roots then multiply out
3. (Original post by Y11_Maths)
Y=k(x-4)(x+5) they are the roots then multiply out
What's with the k?
4. (Original post by RDKGames)
What's with the k?
What my maths teacher told me, in this case k is just 1
5. (Original post by Y11_Maths)
What my maths teacher told me, in this case k is just 1
Ok then, so just leave it out.
6. I have the Equation to Simplify (2*sqrt (3)) - sqrt((7) / ( sqrt(7) - sqrt (3) ) . Does anyone know the answer to this so I can check my calculation of
(4th sqrt (3) * sqrt (7) -1) / (4 ).
7. (Original post by Y11_Maths)
Y=k(x-4)(x+5) they are the roots then multiply out
That's what I would have done, but my teacher was saying something about the symbols switching around so it becomes (x+4)(x-5) = 0 then you multiply it out for x^2 -x - 20 and you get b and c as -1 and 20. But its this switching of the symbols that I don't get - I think I get the rest
8. (Original post by ASJ_12)
That's what I would have done, but my teacher was saying something about the symbols switching around so it becomes (x+4)(x-5) = 0 then you multiply it out for x^2 -x - 20 and you get b and c as -1 and 20. But its this switching of the symbols that I don't get - I think I get the rest
Oh you are correct my bad, I didn’t remember the question properly but my signs are wrong, you are correct. Thanks!
9. (Original post by ASJ_12)
The equation of a graph is of the form y=x^2 +bx+c.
The graph crosses the x- axis at -4 and 5
Sketch the graph and find the values of b and c

I know the answer is c = -20 and b=-1 but I'm not entirely sure how to work it out. Can anyone help
where the graph crosses are the roots of the equation.

Therefore the solutions to the equation y=x^2+bx+c is x=-4 and x= 5

This means that the qudratic eqn is also (x+4)(x-5) = y

expand it.
10. (Original post by Y11_Maths)
Oh you are correct my bad, I didn’t remember the question properly but my signs are wrong, you are correct. Thanks!
So do you know why the symbols switch?
11. (Original post by ASJ_12)
So do you know why the symbols switch?
Yes this is because of the roots. The roots of this quadratic are -4 and 5. Because they are the roots it means the y axis/coordinate will be 0. So if we set is as (x+4)=0 and (x-5)=0 then it means the roots must be x=-4 and x=5
12. (Original post by ASJ_12)
So do you know why the symbols switch?
Either bracket needs to be 0 when you substitute in or . This is only achieved with . Otherwise, has roots 4, -5 instead.
13. (Original post by Y11_Maths)
Yes this is because of the roots. The roots of this quadratic are -4 and 5. Because they are the roots it means the y axis/coordinate will be 0. So if we set is as (x+4)=0 and (x-5)=0 then it means the roots must be x=-4 and x=5
Thanks
14. (Original post by RDKGames)
Either bracket needs to be 0 when you substitute in or . This is only achieved with . Otherwise, has roots 4, -5 instead.
Thanks

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