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completing the square

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    1. Write down the equation of the line of symmetry of the curve y = x2 + 10x + 19.

    2. Hence show that x2 + 2x + 5 is always positive.
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    presuming you want help with this ??
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    (Original post by LoveLifeHate)
    presuming you want help with this ??
    yes please
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    x^2 + 2x + 5 = (x + 1)^2 + 4

    y = (x + 1)^2 + 4

    dy/dx = 2x + 2

    at turning point, 0 = 2x + 2
    x=-1

    at x=-1, y=4

    Because of the shape of the graph, the turning point is always a minimum point, so the lowest point is greater than 0.
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    I always remember HALF, SQUARE, COPY, DONE when completing the square.

    for x^2+bx+c, half b, square b so that you have (x+b/2)^2 - b^2 +c. You should have been told how to interpret a curve in this form.
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    (Original post by WarriorInAWig)
    I always remember HALF, SQUARE, COPY, DONE when completing the square.

    for x^2+bx+c, half b, square b so that you have (x+b/2)^2 - b^2 +c. You should have been told how to interpret a curve in this form.
    Additionally you can expand it out in your head to make sure it matches the original.
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    (Original post by A.J10)
    x^2 + 2x + 5 = (x + 1)^2 + 4

    y = (x + 1)^2 + 4

    dy/dx = 2x + 2

    at turning point, 0 = 2x + 2
    x=-1

    at x=-1, y=4

    Because of the shape of the graph, the turning point is always a minimum point, so the lowest point is greater than 0.
    sorry but how is the minimum point greater than o when x=-1?

    also, q1 the equation is y = x2 + 10x + 19.
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    (Original post by non)
    sorry but how is the minimum point greater than o when x=-1?
    When x=-1, f(x)=4. Since (-1, 4) is the minimum/vertex, there does not exist a point on the curve that has a y-coordinate less than 4. Thus, f(x)≥4, and by extension f(x)=x^2+2x+5>0.

    At a glance, I have no clue how question 2 follows from question 1...
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    (Original post by A.J10)
    x^2 + 2x + 5 = (x + 1)^2 + 4

    y = (x + 1)^2 + 4

    dy/dx = 2x + 2

    at turning point, 0 = 2x + 2
    x=-1

    at x=-1, y=4

    Because of the shape of the graph, the turning point is always a minimum point, so the lowest point is greater than 0.
    You could have stopped at the first line.

    (x+1)^2+4>0
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    (Original post by non)
    1. Write down the equation of the line of symmetry of the curve y = x2 + 10x + 19.
    Given the function

    y = ax^2 + bx + c ,

    completing the square on the RHS tells us that

    Spoiler:
    Show
    y = a\left(x^2 + \dfrac{b}{a}x + \dfrac{c}{a}\right)

    y=a\left(x^2+\dfrac{b}{a}x + 0 + \dfrac{c}{a}\right)

    y=a\left(x^2+\dfrac{b}{a}x + \left(\dfrac{b^2}{4a}-\dfrac{b^2}{4a}\right) + \dfrac{c}{a}\right)

    y=a\left(\left(x^2+\dfrac{b}{a}x  +\dfrac{b^2}{4a^2}\right) + \left(-\dfrac{b^2}{4a^2} + \dfrac{c}{a}\right)\right)

    y=a\left(\left(x+\dfrac{b}{2a} \right)^2 + \left(\dfrac{-b^2+4ac}{4a^2}\right)\right)

    y=a\left(x+\dfrac{b}{2a}\right)^  2 + \dfrac{-b^2+4ac}{4a}



    y=a\left(x-\left(-\dfrac{b}{2a}\right)\right)^2 + \dfrac{-b^2+4ac}{4a}

    Essentially, you have placed the quadratic into vertex form.

    It follows that the x-coordinate of the parabola y=ax+by+c's vertex is always \dfrac{-b}{2a}.

    You should know that the line of symmetry for a parabola is a vertical line (vertical lines have the equation x=?) and that it crosses the vertex (so we know one x value of the equation x=?).

    As such, the equation for said vertical line can be reasoned out.
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    (Original post by non)
    1. Write down the equation of the line of symmetry of the curve y = x2 + 10x + 19.

    2. Hence show that x2 + 2x + 5 is always positive.
    1.

    y is a quadratic so symmetry occurs at its unique turning point.

    y=x^2+10x+19=(x+5)^2-6

    Therefore turning point at (-5,-6)\Rightarrow x=-5 is the line of symmetry of y.

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