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Math Help - Q from From deeqow

  1. #1
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    Q from From deeqow

    Deeqow has PM me this question I am posting it here for anyone who can help.

    Q: The equation of a parabola is given by Y=x2+2x-4 and the straight line by Y=x+2.
    find the point where the parabola crosses x-axis.
    find the point where the parabola crosses y-axis.
    find the parabola axis of symmetry.
    find the coordinates the parabola's minimum point.
    find the point where the parabola crosses the straight line.
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  2. #2
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    Part one

    1. Find the point where Y=x2+2x-4 crosses x axis:

    When y = 0 the parabola is on the x axis so solve 0=x^2+2x-4
    This equation is of the form ax^2 + bx + c
    a=1
    b=2
    c=-4

    To find the roots we use the quadratic equation:

    plug and chug and you get -1+sqrt(5) and -1-sqrt(5)
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  3. #3
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    part 2

    2. To find the point where Y=x^2+2x-4 crosses the y axis we need to find the solution to the equation when x = 0:

    Y=0^2+2*0-4
    Y=-4
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  4. #4
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    Here is the graph
    Attached Thumbnails Attached Thumbnails Q from From deeqow-graph.gif  
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  5. #5
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    To find the parabola's axis of symmetry and the minumum point you need to either use a graphing calculator to find the min on the curve or take the derivative and set it equal to zero because the slope at the min is always zero.

    Derivative of x^2 + 2x - 4 = 2x+2

    2x +2 = 0

    x=-1 this is the minimum point

    The parabola's axis of symmetry is a vertical line through this point which can be expressed as x=-1
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  6. #6
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    The axis of symmetry can be found more easily by x=-b/2a, the vertex or minimum point is obviously (-b/2a,f(-b/2a)), the points of intersection can be found by using the substitution method and then factoring.
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  7. #7
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    y = x^2 +2x -4 ....(1)
    Because it is the x that is squared, and the x^2 is positive, then this is a vertical parabola that opens upward.

    A standard form of a vertical parabola that opens upward is
    (y-k) = a(x-h)^2 ....(2)
    where
    (h,k) is the vertex or lowest point.
    a is any real number.

    Let us transform (1) into (2).
    y = x^2 +2x -4 ....(1)
    We complete the square of the x terms,
    y +4 = x^2 +2x
    y +4 = x^2 +2x +(2/2)^2 -(2/2)^2
    y +4 = x^2 +2x +1 -1
    y +4 +1 = x^2 +2x +1
    y +5 = (x+1)^2 ....(3)
    There, the vertical parabola in standard form.

    That means k = -5; h = -1, a=1
    Or the lowest point or vertex is (h,k) = (-1,-5).

    -------------------
    >>>find the axis of symmetry.

    The axis of symmetry of a vertical parabola is a vertical line that passes through the vertex.
    Therefore, the axis of symmetry is the line x = -1. ....answer.

    ------------
    >>>Find the coordinates of the parabola's minimum point.

    The mimimum point of a vertical parabola is its lowest point or vertex.
    Therefore, the minimum point is (-1,-5). ....answer.

    -------------
    >>>find the point where the parabola crosses x-axis.

    When the parabola crosses the x-axis, the y-value is zero, or y=0.
    So,
    y +5 = (x+1)^2 ....(3)
    0 +5 = (x+1)^2
    (x+1)^2 = 5
    Take the square roots of both sides,
    (x+1) = +,-sqrt(5)
    Or,
    x = -1 +sqrt(5) ,or about 1.236
    x = -1 -sqrt(5) , or about -3.236

    That means there are 2 points where the parabola crossses the x-axis:
    (-3.236,0) and (1.236,0) ....answer.

    ------------
    >>>find the point where the parabola crosses y-axis.

    When the parabola crosses the y-axis, the x-value is zero, or x=0.
    So,
    y +5 = (x+1)^2 ....(3)
    y +5 = (0+1)^2
    y +5 = (1)^2
    y +5 = 1
    y = 1 -5
    y = -4

    That means the parabola crossses the y-axis at only one point:
    (0,-4) ....answer.

    ---------------
    >>>find the point where the parabola crosses the straight line.

    When the parabola crosses the the straight line y = x+2, at the points of intersection the parabola and the straight line have the same x- and y-values.
    So,
    y = x+2
    Substitute that in (1),
    y = x^2 +2x -4 ....(1)
    x+2 = x^2 +2x -4
    0 = x^2 +2x -x -4 -2
    0 = x^2 +x -6
    Or,
    x^2 +x -6 = 0
    Factor that,
    (x+3)(x-2) = 0

    So,
    x+3 = 0
    x = -3

    x-2 = 0
    x = 2

    When x = -3,
    y = x+2 = -3+2 = -1
    That is point (-3,-1)

    when x = 2,
    y = x+2 = 2+2 = 4
    That is point (2,4)

    Therefore, the the parabola crosses the straight at points (-3,-1) and (2,4). ....answer.
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