Results 1 to 4 of 4

Math Help - Finding points of intersection

  1. #1
    Newbie
    Joined
    May 2009
    Posts
    1

    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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    No one in Particular VonNemo19's Avatar
    Joined
    Apr 2009
    From
    Detroit, MI
    Posts
    1,823
    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?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570

    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
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,807
    Thanks
    116
    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
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 4
    Last Post: January 19th 2010, 08:35 AM
  2. Finding points of intersection (polar graphs)
    Posted in the Calculus Forum
    Replies: 6
    Last Post: December 16th 2009, 02:38 AM
  3. Replies: 1
    Last Post: February 16th 2009, 03:27 PM
  4. help finding x-coordinates of points of intersection
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: December 1st 2008, 03:15 PM
  5. Points of intersection
    Posted in the Calculus Forum
    Replies: 3
    Last Post: January 30th 2008, 06:03 PM

Search Tags


/mathhelpforum @mathhelpforum