# maths question - quadratic graphs

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#1
can someone help me with this question please??
0
1 year ago
#2
(Original post by sqrt of 5)
can someone help me with this question please??
You have three points the function and its derivative passes through, so you have 3 linear equations in three unknowns (a,b,c) to solve.
0
1 year ago
#3
(Original post by mqb2766)
You have three points the function and its derivative passes through, so you have 3 linear equations in three unknowns (a,b,c) to solve.
Or, you can start by reasoning that if the function has one root at x = 4 and its minimum is at x = 5, then, by symmetry, the other root must be at x = ....
2
#4
(Original post by old_engineer)
Or, you can start by reasoning that if the function has one root at x = 4 and its minimum is at x = 5, then, by symmetry, the other root must be at x = ....
(6,0)? now i can sub -4 and -6 and the solve it right?
0
1 year ago
#5
(Original post by sqrt of 5)
(6,0)? now i can sub -4 and -6 and the solve it right?
Or write the quadratic down in terms of its factors (roots) and expand (with a multiplier on the front to get the value of the minimum right).
Last edited by mqb2766; 1 year ago
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#6
(Original post by mqb2766)
Or write the quadratic down in terms of its factors (roots) and expand (with a multiplier on the front to get the value of the minimum right).
but i only have 4 what's the other root?
0
1 year ago
#7
(Original post by sqrt of 5)
can someone help me with this question please??
If it has a minimum point of (5,-3) and if it passes through (4,0) then this means that it crosses the x axis at 4.
We can work this out by completing the square backwards, and then by using the factor theorem!

y = k(x-5)^2 - 3

y = k(x-5)(x-5) - 3

y = k(x^2-10x+25) - 3

y = kx^2 - 10kx + 25k - 3

We know that x - 4 is a factor so by factor theorem:

f(4) = 16k - 40k + 25k - 3 = 0

f(4) = k - 3 = 0

Therefore k = 3

Substitute the k value back into the original equation:

y = 3x^2 - 30x + 75 - 3

y = 3x^2 - 30x + 72

(And so the second root is 6) Any questions, feel free to ask!
Last edited by dxnixl; 1 year ago
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#8
(Original post by dxnixl)
If it has a minimum point of (5,-3) and if it passes through (4,0) then this means that it crosses the x axis at 4.
We can work this out by completing the square backwards, and then by using the factor theorem!

y = k(x-5)^2 - 3

y = k(x-5)(x-5) - 3

y = k(x^2-10x+25) - 3

y = kx^2 - 10kx + 25k - 3

idek what the factor theorem is lol
0
1 year ago
#9
(Original post by sqrt of 5)
but i only have 4 what's the other root?
The roots are symmetric about the minimum so the other one is 6.
x = 4, 6
so the factors are
(x-4), (x-6)
If a = 1, that would give the quadratic when you expand them? So
a(x-4)(x-6)
is the quadratic. Use the minimum point to determine "a".

(Original post by dxnixl)
If it has a minimum point of (5,-3) and if it passes through (4,0) then this means that it crosses the x axis at 4.
We can work this out by completing the square backwards, and then by using the factor theorem!
Hints not solutions
0
1 year ago
#10
(Original post by sqrt of 5)
idek what the factor theorem is lol
Oh lol... you don't really need to know what it means.

Basically, if you know one root and have an equation (as we created with the value k), you substitute the root you know into it and make it equal to 0. When x is 4, y is 0. If we know this, we can work out what k is. After that, we just substitute k back into the equation.
Last edited by dxnixl; 1 year ago
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#11
(Original post by mqb2766)
The roots are symmetric about the minimum so the other one is 6.
x = 4, 6
so the factors are
(x-4), (x-6)
If a = 1, that would give the quadratic when you expand them? So
a(x-4)(x-6)
is the quadratic. Use the minimum point to determine "a".

Hints not solutions
(Original post by sqrt of 5)
(6,0)? now i can sub -4 and -6 and the solve it right?
Yea thats what i was trying to say b4

(Original post by dxnixl)
Oh lol... you don't really need to know what it means.

Basically, if you know one root and have an equation (as we created with the value k), you substitute the root you know into it and make it equal to 0. When x is 4, y is 0. If we know this, we can work out what k is. After that, we just substitute k back into the equation.
thank you but im gonna stick to the easier version 😭 i still dont get your method. am i supposed to know the factor theorem?
0
1 year ago
#12
(Original post by sqrt of 5)
Yea thats what i was trying to say b4

thank you but im gonna stick to the easier version 😭 i still dont get your method. am i supposed to know the factor theorem?
It’s super easy dw. If you haven’t learnt it then you either will or don’t need to. Let me go through it again in a more easy way:

We expanded and got this as our equation:
y = kx^2 - 10kx + 25k - 3

We know that when x = 4, y = 0 (4,0) so we substitute 4 into x, and 0 into y:

0= 16k - 40k + 25k - 3

0 = k - 3

Therefore k = 3

And after we know what k is, we substitute it back into the equation we made at the beginning 0
#13
(Original post by dxnixl)
It’s super easy dw. If you haven’t learnt it then you either will or don’t need to. Let me go through it again in a more easy way:

We expanded and got this as our equation:
y = kx^2 - 10kx + 25k - 3

We know that when x = 4, y = 0 (4,0) so we substitute 4 into x, and 0 into y:

0= 16k - 40k + 25k - 3

0 = k - 3

Therefore k = 3

And after we know what k is, we substitute it back into the equation we made at the beginning oh i get it
do u know where i can use this theorem? cuz i dont i will remember it i need to do some practice
0
1 year ago
#14
(Original post by sqrt of 5)
oh i get it
do u know where i can use this theorem? cuz i dont i will remember it i need to do some practice
You use it in mostly questions like these. There might be a few harder questions but if you’ve got to find a quadratic expression, then it’ll come in useful.

Just remember it for questions like these that need you to work out the equation of a graph. I don’t really know how else you could use it 0
#15
(Original post by dxnixl)
You use it in mostly questions like these. There might be a few harder questions but if you’ve got to find a quadratic expression, then it’ll come in useful.

Just remember it for questions like these that need you to work out the equation of a graph. I don’t really know how else you could use it ok thank u :3
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