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    why does putting an equation into completed square form show the vertex of the graph?

    So for example:
    f(x) = x2 + 4x + 3
    In completed square form is x = -2 +- root1 (I think)
    And from this the vertex is (-2,-1)

    But why?
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    (Original post by surina16)
    why does putting an equation into completed square form show the vertex of the graph?

    So for example:
    f(x) = x2 + 4x + 3
    In completed square form is x = -2 +- root1 (I think)
    And from this the vertex is (-2,-1)

    But why?
    I am afraid you didn't complete the square properly.
    the ans is ((x+2)^2) -1
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    As Azhar said, the square wasn't completed properly. But imagine you had the graph y = f(x) where f(x) = x2

    Have you learnt graph translations?

    What if we had y = f(x+2)? This means f(x+2) = (x+2)2 !

    What does f(x+2) mean from the f(x) graph? It means that the graph is shifted 2 units to the left.

    So what if we plonk on your -1?

    y = f(x+2) - 1

    this means we get y = (x+2)2 - 1, which happens to be your graph.

    What does a translation of y = f(x+2) - 1 represent? It represents shifting to the left 2 units, as we saw earlier. But NOW we have this -1 outside. So what happens if we take away 1 from all f(x+2) values? The graph shifts down by 1 unit.

    So in your example, the y = x2 graph shifted 2 units left and 1 unit down where the original vertex was the origin. What happens to the point (0, 0) when shifted 2 left and 1 down? It becomes (-2, -1)
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    (Original post by surina16)
    why does putting an equation into completed square form show the vertex of the graph?

    So for example:
    f(x) = x2 + 4x + 3
    In completed square form is x = -2 +- root1 (I think)
    And from this the vertex is (-2,-1)

    But why?
    I'll complete the square correctly for you: x^2 + 4x + 3 = (x+2)^2 - 1.

    Now, since you know that square quantities are always \geq 0 then the minimum value of this expression occurs when the squared term is zero.

    This means that \min(x^2 + 4x + 3) = \min((x+2)^2 - 1) occurs precisely when (x+2)^2 = 0 \Rightarrow x = -2.

    The minimum value of function at the point x=-2 is \min (x^2 + 4x + 3) = \min((x+2)^2 -1 ) = 0-1 = -1.
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    (Original post by Azhar Rana)
    I am afraid you didn't complete the square properly.
    the ans is ((x+2)^2) -1
    How did you get that? this is what I did:

    x^2 + 4x + 3 = 0
    (x+2)^2 -4 + 3 = 0
    (x+2)^2 - 1 = 0
    (x+2)^2 = 1
    x+2 = +-root 1
    x= -2 +- root 1

    which bit was wrong?
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    (Original post by surina16)
    How did you get that? this is what I did:

    x^2 + 4x + 3 = 0
    (x+2)^2 -4 + 3 = 0
    (x+2)^2 - 1 = 0
    (x+2)^2 = 1
    x+2 = +-root 1
    x= -2 +- root 1

    which bit was wrong?
    That's not the vertex, that's you finding the roots of the quadratic. Oh, and \sqrt{1} = 1.

    Also: moved to maths.
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    (Original post by Student403)
    ...
    Damn, you win this one. At least our explanations are substantially different to warrant the ninja.
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    (Original post by Azhar Rana)
    I am afraid you didn't complete the square properly.the ans is ((x+2)^2) -1
    (Original post by Student403)
    As Azhar said, the square wasn't completed properly. But imagine you had the graph y = f(x) where f(x) = x2

    Have you learnt graph translations?

    What if we had y = f(x+2)? This means f(x+2) = (x+2)2 !

    What does f(x+2) mean from the f(x) graph? It means that the graph is shifted 2 units to the left.

    So what if we plonk on your -1?

    y = f(x+2) - 1

    this means we get y = (x+2)2 - 1, which happens to be your graph.

    What does a translation of y = f(x+2) - 1 represent? It represents shifting to the left 2 units, as we saw earlier. But NOW we have this -1 outside. So what happens if we take away 1 from all f(x+2) values? The graph shifts down by 1 unit.

    So in your example, the y = x2 graph shifted 2 units left and 1 unit down where the original vertex was the origin. What happens to the point (0, 0) when shifted 2 left and 1 down? It becomes (-2, -1)
    (Original post by Zacken)
    I'll complete the square correctly for you: x^2 + 4x + 3 = (x+2)^2 - 1.

    Now, since you know that square quantities are always \geq 0 then the minimum value of this expression occurs when the squared term is zero.

    This means that \min(x^2 + 4x + 3) = \min((x+2)^2 - 1) occurs precisely when (x+2)^2 = 0 \Rightarrow x = -2.

    The minimum value of function at the point x=-2 is \min (x^2 + 4x + 3) = \min((x+2)^2 -1 ) = 0-1 = -1.
    Ah, that makes sense! Thank you everyone
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    (Original post by Zacken)
    Damn, you win this one. At least our explanations are substantially different to warrant the ninja.
    7.5/11 :rofl:

    And yeah! I think yours is a lot simpler though :lol: And ofc in maths simpler = better
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    (Original post by Student403)
    7.5/11 :rofl:

    And yeah! I think yours is a lot simpler though :lol: And ofc in maths simpler = better
    Daaamn, getting there. K : D ratio :rofl:

    Nah, I think yours is just as nice (if not nicer), transformations can be a very elegant thing!
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    (Original post by Zacken)
    Daaamn, getting there. K : D ratio :rofl:

    Nah, I think yours is just as nice (if not nicer), transformations can be a very elegant thing!
    K/D omg lol

    Thanks! I do love translations because they seem so intuitive to me :dontknow:
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    (Original post by Zacken)
    That's not the vertex, that's you finding the roots of the quadratic. Oh, and \sqrt{1} = 1.

    Also: moved to maths.
    Oops Thanks for your help
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    (Original post by Zacken)
    Damn, you win this one. At least our explanations are substantially different to warrant the ninja.
    lol, you guys
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    (Original post by surina16)
    Oops Thanks for your help
    Remember: f(x) = ax^2 + bx + c has the vertex at \displaystyle \left(-\frac{b}{2a}, f\left(-\frac{b}{2a}\right)\right) but the roots at \displaystyle x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} - they are two different things.

    No problem.
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    (Original post by surina16)
    How did you get that? this is what I did:

    x^2 + 4x + 3 = 0
    (x+2)^2 -4 + 3 = 0
    (x+2)^2 - 1 = 0
    (x+2)^2 = 1
    x+2 = +-root 1
    x= -2 +- root 1

    which bit was wrong?
    remember,
    y=x^2 the vertex is (0,0)
    applying transformation
    x+2 shifts x to the -ve x axis by 2 units.
    y=(x)^2 -1
    here we shif the Y by 1 in the -ve y axis.
    so, the result is (x-2)^2 -1 the vertex is now -2,1
    the vertex is shifted in the -x axis by 2 units and 1 unit in the -y direction,

    hence the vertex is (-2,-1)
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    (Original post by Azhar Rana)
    lol, you guys
    :hat2:
 
 
 
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