# maths question - quadratic graphs

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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.

**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.

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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 = ....

**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 = ....

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(6,0)? now i can sub -4 and -6 and the solve it right?

**sqrt of 5**)(6,0)? now i can sub -4 and -6 and the solve it right?

Last edited by mqb2766; 1 year ago

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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).

**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).

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#7

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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(Original post by

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!

**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**
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(Original post by

but i only have 4 what's the other root?

**sqrt of 5**)but i only have 4 what's the other root?

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

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!

**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!

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#10

(Original post by

idek what the factor theorem is lol

**sqrt of 5**)idek what the factor theorem is lol

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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(Original post by

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

**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

(6,0)? now i can sub -4 and -6 and the solve it right?

**sqrt of 5**)(6,0)? now i can sub -4 and -6 and the solve it right?

(Original post by

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.

**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.

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#12

(Original post by

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?

**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?

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

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(Original post by

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:

We know that when

Therefore

And after we know what k is, we substitute it back into the equation we made at the beginning

**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

do u know where i can use this theorem? cuz i dont i will remember it i need to do some practice

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#14

(Original post by

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

**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

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

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(Original post by

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

**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

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