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Math Help - Finding points of intersection

  1. #1
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    Finding points of intersection

    Find the points of intersection of the graphs in the system.

    x^2= 6y
    y= -x

    Please show me step by step how to solve this. The answers in the back of the book are

    (0,0) and (-6,6).

    EDIT:

    Help also appreciated with the following problem:

    x^2 + y^2 = 20
    y= x-4

    Find points of intersection.
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  2. #2
    No one in Particular VonNemo19's Avatar
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    All youve gotta do is solve for x or y and then substitute

    watch

    x^2=6y\Longleftrightarrow{y=\frac{x^2}{6}}

    Now

    since y=y

    \frac{x^2}{6}=-x\Longleftrightarrow{x^2=-6x}\Longleftrightarrow{x=-6}

    How's that?
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  3. #3
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    Points of intersection

    Hello cuandoyando
    Quote Originally Posted by cuandoyando View Post
    Find the points of intersection of the graphs in the system.

    x^2= 6y
    y= -x

    Please show me step by step how to solve this. The answers in the back of the book are

    (0,0) and (-6,6).
    A point on any graph is represented by a value of x and a value of y which are connected by the equation of the graph - in other words these values of x and y 'satisfy' the equation of the graph. We can find where two graphs intersect if we can find a value of x and a value of y which satisfy the equations of both graphs at the same time. The two equations you have are:

    x^2 = 6y and y = -x

    So we need to find values of x and y that satisfy both these equations at the same time.

    Now if y = -x, then 6y = -6x. So if, at the same time, 6y = x^2, then we must have:

    x^2 = -6x

    \Rightarrow x^2+6x=0

    \Rightarrow x(x+6) = 0

    \Rightarrow x = 0 or -6

    So, bearing in mind that y = -x:

    When x = 0, y = 0, and when x = -6, y = 6.

    So the points where the graphs intersect are (0,0) and (-6,6).


    Help also appreciated with the following problem:

    x^2 + y^2 = 20
    y= x-4

    Find points of intersection.
    Use the same technique here. If y = (x-4), then we can replace y by (x-4) in the equation x^2 + y^2=20 (in other words, substitute for y) to get:

    x^2 + (x-4)^2 = 20

    Can you continue? (You'll need to expand the brackets, simplify and the solve the quadratic equation by formula.)

    Grandad
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  4. #4
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    Quote Originally Posted by VonNemo19 View Post
    All youve gotta do is solve for x or y and then substitute

    watch

    x^2=6y\Longleftrightarrow{y=\frac{x^2}{6}}

    Now

    since y=y

    \frac{x^2}{6}=-x\Longleftrightarrow{x^2=-6x}\Longleftrightarrow{x=-6}

    How's that?
    I don't want to pick at you but you removed one solution by dividing by x:

    \frac{x^2}{6}=-x\Longleftrightarrow{x^2+6x=0}\Longleftrightarrow{  x(x+6)=0}

    A product equals zero if one factor equals zero. Therefore:

    x = 0~\vee~x=-6
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