Results 1 to 7 of 7

Math Help - exam help

  1. #1
    Newbie
    Joined
    Mar 2010
    Posts
    12

    exam help

    um could anyone plz help me with this queston thats on an exam tommorow....and i need at least a B+....


    y= 4√(x+4) -2
    y= ⅔x +7

    simultaneous equation....i need to sketch a graph, label intercepts and state their value

    thx in advance....
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member Awsom Guy's Avatar
    Joined
    Jun 2009
    From
    Mars
    Posts
    186
    have you tried to graph that on a calculator or something.

    The second one is a straight line

    Quote Originally Posted by Awsom Guy View Post
    The second one is a straight line

    the intercepts for the second equation is y is 7 and x is 10. I just graphed this into my calculator so it is right and the other equation is crazy I am still trying to work it out.
    Last edited by mr fantastic; March 15th 2010 at 01:40 AM. Reason: Merged posts
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member Awsom Guy's Avatar
    Joined
    Jun 2009
    From
    Mars
    Posts
    186
    If I say this could you do it:
    4sqrt(x+4) - 2 (1)
    2/3x + 7 (2)

    Make equation 1 like this solve it first:
    4(x^1/2)+2) -2
    4x^1/2 + 8 -2
    4(x^1/2) + 6

    Therefore we get our 2 equations:
    4(x^1/2) + 6
    (2/3)x + 7
    Can you solve from there...
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Mar 2010
    Posts
    12
    Quote Originally Posted by Awsom Guy View Post
    If I say this could you do it:
    4sqrt(x+4) - 2 (1)
    2/3x + 7 (2)

    Make equation 1 like this solve it first:
    4(x^1/2)+2) -2
    4x^1/2 + 8 -2
    4(x^1/2) + 6

    Therefore we get our 2 equations:
    4(x^1/2) + 6
    (2/3)x + 7
    Can you solve from there...

    ummm is "^" square root sign?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member Awsom Guy's Avatar
    Joined
    Jun 2009
    From
    Mars
    Posts
    186
    yes like 2^2=4
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member Bacterius's Avatar
    Joined
    Nov 2009
    From
    Wellington
    Posts
    927
    1) Sketch a graph : just plug in x = -3, -2, -1, 0, 1, 2, 3 and get the corresponding y values. Then ... plot. Note that for the second one only two points are required since it is the equation of a line.

    2) Label intercepts : that shouldn't be too hard, for the x-intercepts just set y = 0 and solve for x in both cases, for the y-intercepts just set x = 0 and solve for y in both cases. You therefore get their values which you can then sketch.


    As you mentioned "simultaneous equations" I assume you need to solve the system at some point (know where the two curves intersect). You can apply the following method :

    Just set 4 \sqrt{x + 4} - 2 = \frac{2}{3} x + 7 and solve for x :

    4 \sqrt{x + 4} - 2 = \frac{2}{3} x + 7

    4 \sqrt{x + 4} = \frac{2}{3} x + 9

    \left ( 4 \sqrt{x + 4} \right ) ^2 = \left ( \frac{2}{3} x + 9 \right ) ^2

    16(x + 4) = \left ( \frac{2}{3} x \right )^2 + \frac{36}{3} x + 9^2

    16x + 64 = \frac{4}{9} x^2 + 12x + 81

    0 = \frac{4}{9} x^2 - 4x + 17

    9 \times 0 = 9 \left ( \frac{4}{9} x^2 - 4x + 17 \right )

    0 = 4 x^2 - 36x + 153

    4 x^2 - 36x + 153 = 0

    Calculate the discriminant :

    \Delta = b^2 - 4ac = (-36)^2 + 4 \times 4 \times 153 = -1152

    It is negative therefore this equation has no real solution for x, thus the two curves (the curve and the line) never meet (never intersect).

    Does it make sense ?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Mar 2010
    Posts
    12
    Quote Originally Posted by Awsom Guy View Post
    yes like 2^2=4
    thank u so much yet again then!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. 2nd PDE for Exam
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: September 13th 2010, 03:43 PM
  2. PDE for Exam
    Posted in the Differential Equations Forum
    Replies: 4
    Last Post: September 13th 2010, 11:49 AM
  3. Help for exam
    Posted in the Statistics Forum
    Replies: 0
    Last Post: June 8th 2010, 08:41 AM
  4. exam help
    Posted in the Algebra Forum
    Replies: 1
    Last Post: December 13th 2009, 10:33 AM
  5. Help me with log. tmr exam!
    Posted in the Algebra Forum
    Replies: 1
    Last Post: March 25th 2009, 03:54 AM

Search Tags


/mathhelpforum @mathhelpforum