Page 2 of 2 FirstFirst 12
Results 16 to 21 of 21

Math Help - Equations and Graph

  1. #16
    Senior Member ecMathGeek's Avatar
    Joined
    Mar 2007
    Posts
    436
    Quote Originally Posted by GAdams View Post
    I got the first equation by using the graph. I plotted -1 1/2 on the x axis and -1/2 on the y axis and drew the line. I then labelled it y=x+1.

    I can't work out how you got the second onethough.
    Don't plot x=-1.5 and then plot y=-0.5. You're supposed to find the point (-1.5,-0.5) on the graph. When you find that point (on the graph provided), you will see that two of the lines on that graph intersect that point. The two lines that are intersecting at that point are the answers for the problem.
    Follow Math Help Forum on Facebook and Google+

  2. #17
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by GAdams View Post
    I got the first equation by using the graph. I plotted -1 1/2 on the x axis and -1/2 on the y axis and drew the line. I then labelled it y=x+1.
    that is incorrect. not only is your method off (as you don't have to draw anything for these questions) but the line you drew would not be y = x + 1

    ecMathGeek gave a short explanation for this question, look at it


    I can't work out how you got the second onethough.
    They gave the solutions as -3.3 or 0.3 so i looked at the graph for those x-values to see if any graphs intersected at that those points. i found that the quadratic and the line y = -x - 2 intersected there, so i just equated them (since intersecting points occur when one graph equals the other) and simplified the answer to look like a quadratic. when i used the quadratic formula, i was just verifying my answer, you would not have to do that.......do you understand now, or is something still unclear?
    Follow Math Help Forum on Facebook and Google+

  3. #18
    Member GAdams's Avatar
    Joined
    Apr 2007
    Posts
    171
    Awards
    1
    OK. Here goes:

    (a) y=x+1 and y=-x-2
    Got it. Thanks.

    (b) x = -3.3 or 0.3
    Looked at the graph, found that the following two lines interestected at those points. So they equal each other at those points:

    y= x^2 + 2x -3 and y= -x -2

    x^2 + 2x -3 = -x -2

    x^2 + 3x -1 = 0

    Which is the answer given by you guys.

    The answers with the paper give a mark for this, but the final answers is given as:

    x^2 + 2x - 3 = x + 1

    -How did they get that?

    (c) I read what you wrote, but still cant see how to give the solutions from the graph.
    Follow Math Help Forum on Facebook and Google+

  4. #19
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by GAdams View Post
    OK. Here goes:

    (a) y=x+1 and y=-x-2
    Got it. Thanks.
    good


    (b) x = -3.3 or 0.3
    Looked at the graph, found that the following two lines interestected at those points. So they equal each other at those points:

    y= x^2 + 2x -3 and y= -x -2

    x^2 + 2x -3 = -x -2

    x^2 + 3x -1 = 0

    Which is the answer given by you guys.

    The answers with the paper give a mark for this, but the final answers is given as:

    x^2 + 2x - 3 = x + 1

    -How did they get that?
    i think you are looking at the wrong question, that's what we used for part (c). if you check the solutions to that equation, you will realize it is not -3.3 or 0.3


    (c) I read what you wrote, but still cant see how to give the solutions from the graph.
    we realize that we can construct the given quadratic using two of the original graphs given

    we can take y = x^2 + 2x - 3 and y = x + 1 and subtract them, the result will be the given quadratic.

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

    now we want to solve x^2 + x - 4 = 0

    that means we want x^2 + 2x - 3 - (x + 1) = 0

    but that's the same as x^2 + 2x - 3 = x + 1

    and so the answers will be where the graphs y = x^2 + 2x - 3 and y = x + 1 intersect
    Follow Math Help Forum on Facebook and Google+

  5. #20
    Member GAdams's Avatar
    Joined
    Apr 2007
    Posts
    171
    Awards
    1
    The final answer for (b) in the exam paper is definitely what I posted. Maybe it's a misprint and should have been with the next question.

    Anyway, You said that we can construct the given quadratic using two of the lines from the graph. How would you know which ones to use and whether to subtract or add? Trial and error?
    Follow Math Help Forum on Facebook and Google+

  6. #21
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by GAdams View Post
    The final answer for (b) in the exam paper is definitely what I posted. Maybe it's a misprint and should have been with the next question.

    Anyway, You said that we can construct the given quadratic using two of the lines from the graph. How would you know which ones to use and whether to subtract or add? Trial and error?
    pretty much, but in this case it was easy. in the given quadratic, the constant term was -4. we know we had to combine the original quadraic with one of the lines, so choosing the line with constant term + 1 was the only choice that made sense, and we know we had to subtract because -3 - 1 = -4
    Follow Math Help Forum on Facebook and Google+

Page 2 of 2 FirstFirst 12

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: February 16th 2011, 09:41 PM
  2. graph equations
    Posted in the Algebra Forum
    Replies: 4
    Last Post: February 21st 2010, 01:50 AM
  3. Can someone show me a graph of two equations?
    Posted in the Calculus Forum
    Replies: 1
    Last Post: September 20th 2009, 10:52 PM
  4. equations that do not graph correctly
    Posted in the Calculus Forum
    Replies: 3
    Last Post: March 19th 2009, 04:08 PM
  5. Solving equations - to Graph!
    Posted in the Math Topics Forum
    Replies: 2
    Last Post: November 4th 2008, 04:46 PM

Search Tags


/mathhelpforum @mathhelpforum