Find the equation of a curve with a given turning point

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dthomas86
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#1
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#1
Hey everyone, I've been tearing my hair out with this question I'm really not sure what you are meant to do and my book has not explained it anywhere how to solve such a question.

Question: Each curve shown is quadratic of the form y = x^2 + bx + c and the turning point is shown (in this case the turning point is (2,3) ). Find the equation of each curve.

Possible answer: Do I substitute in the value of x and y (in this case x=2, y=3) into the equation y = x^2 +bx + c to get 3 = 2^ + 2b + c (although I thought c would be 3 as this is the point where the curve crosses the y-axis? So I would get y= 2^2 + 2b + 3).

Any further help I'll greatly appreciate it. Cheers everyone!!
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Indeterminate
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#2
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#2
(Original post by dthomas86)
Hey everyone, I've been tearing my hair out with this question I'm really not sure what you are meant to do and my book has not explained it anywhere how to solve such a question.

Question: Each curve shown is quadratic of the form y = x^2 + bx + c and the turning point is shown (in this case the turning point is (2,3) ). Find the equation of each curve.

Possible answer: Do I substitute in the value of x and y (in this case x=2, y=3) into the equation y = x^2 +bx + c to get 3 = 2^ + 2b + c (although I thought c would be 3 as this is the point where the curve crosses the y-axis? So I would get y= 2^2 + 2b + 3).

Any further help I'll greatly appreciate it. Cheers everyone!!
Hint:

If

f(x) = (x+a)^2 + b

then

(-a,b)

is the turning point
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dthomas86
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#3
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#3
So do I need to complete the square for y = x^2 + bx + c???


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dthomas86
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#4
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#4
Plus what has happened to c?


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Indeterminate
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#5
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#5
(Original post by dthomas86)
So do I need to complete the square for y = x^2 + bx + c???


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No.

In your case,

(-a,b) = (2,3)

Should be easy now
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HenryD
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#6
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#6
Hint: The gradient at each of this point on all the graphs will be 0. So dy/dx=0. If you've done differentiation then you should be able to go from there
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dthomas86
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#7
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#7
Sorry guys I'm being dull I haven't really come across differentiation yet as its quite early on in the book.

The answer in the book is y = x^2 - 4x + 7

I still don't know how to get this answer.


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Indeterminate
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#8
(Original post by dthomas86)
Sorry guys I'm being dull I haven't really come across differentiation yet as its quite early on in the book.

The answer in the book is y = x^2 - 4x + 7

I still don't know how to get this answer.


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You don't need differentiation.

Note that (a,b)

is the turning point of the equation

f(x)=(x-a)^2 + b

What are a and b in your case?

Now expand the brackets
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Albino
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#9
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#9
(Original post by dthomas86)
Sorry guys I'm being dull I haven't really come across differentiation yet as its quite early on in the book.

The answer in the book is y = x^2 - 4x + 7

I still don't know how to get this answer.


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Differentiate your f(x). Sub the x value you have into the equation, rearrange to find b.
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HenryD
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#10
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#10
(Original post by dthomas86)
Sorry guys I'm being dull I haven't really come across differentiation yet as its quite early on in the book.

The answer in the book is y = x^2 - 4x + 7

I still don't know how to get this answer.


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It's not the only way of doing it as indeterminate says but I always use it and most of the people I know prefer it too. Have you come across completing the square? That's the other easy way to do it.
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dthomas86
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#11
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#11
How did you get the equation (x - a)^2 + b^2 in the first place? Where did the coefficient a come from and what happened to c?

Sorry guys its been a long day and it hasn't been clearly explained in the book how to tackle such a question. Cheers.


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ecw412
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#12
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#12
(Original post by dthomas86)
How did you get the equation (x - a)^2 + b^2 in the first place? Where did the coefficient a come from and what happened to c?

Sorry guys its been a long day and it hasn't been clearly explained in the book how to tackle such a question. Cheers.


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The a and b in the equation you just quoted are not the same a and b as in your original equation, they come from the coordinates of the turning point, as explained by someone above.
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dthomas86
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#13
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#13
Yeah I know how to complete the square would I get: (x + b/2 )^2 - (b/2)^2 + c if I completed the square for the quadratic y = x^2 + bx + c


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FO12DY
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#14
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#14
\frac{d}{dx}|_{x=2} (x^2 + bx + c) = 0

2(2) + b = 0 

   

\Rightarrow    b = -4 

   

\Rightarrow    c = (3) - (2)^2 + 4(2) = 7
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dthomas86
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#15
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#15
Without using differentiation could you show me how to solve it by completing the square? Cheers


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FO12DY
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#16
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#16
(Original post by dthomas86)
Without using differentiation could you show me how to solve it by completing the square?
Is it on a GCSE paper?!

/if not, why do you want to avoid differentiation?
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dthomas86
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#17
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#17
Because the chapter in the book I have got is about completing the square.

I'm not sure why a question like that is in the book so early on. As I haven't touched on differentiation yet.

Do you know how to solve it by completing the square?


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Mark85
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#18
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#18
(Original post by dthomas86)
Yeah I know how to complete the square would I get: (x + b/2 )^2 - (b/2)^2 + c if I completed the square for the quadratic y = x^2 + bx + c


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The point is that completing the square shows you that the turning point in y = x^2 + bx + c is at x=-b/2
so if you know the turning point, you know what -b/2 is. Use the other coordinate of the turning point to find c
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Dualcore
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#19
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#19
(Original post by dthomas86)
Because the chapter in the book I have got is about completing the square.

I'm not sure why a question like that is in the book so early on. As I haven't touched on differentiation yet.

Do you know how to solve it by completing the square?


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so you have x^2+bx+c=y, using completing the square you get
y=(x+b/2)^2-(b/2)^2+c
As you can see apart from thee the bit in the brackets the other terms are constants in terms of b and c.
You know the turning point is at (2,3), if you think about what the graph looks like the minimum point is where the y is the smallest value it can be and that is where x=2, as anything squared is positive the value of the first term must be zero so ((2)+(b/2))^2=0 therefore ((2)+(b/2))=0, and now we consider the y value at this point, it is 3 so we can say that 3=-(b/2)^2+c, we can use the value of b we calculated before to work this out.

Hope this helped
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FO12DY
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#20
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#20
I only ask because by the time you've covered all the material, I'd wager on you solving it by differentiation without batting an eyelid, without considering completing the square.
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