Results 1 to 12 of 12

Math Help - Functions help (test soon)

  1. #1
    Junior Member
    Joined
    Dec 2007
    Posts
    41

    Functions help (test soon)

    Hey guys, i have a test soon and just wanted to get a few things cleared up. Thanks

    1. a. How do u identify the x-intercepts of the graph of y=x(x^2-9a^2) (note: a>0)

    b. How would u go about sketching the graph of y= x(x^2-9a^2)

    2. If the equation 2x^2 + 6=ax has exactly one solution, find the value(s) of a, such that this is true.

    3. Is it possible to find the x intercepts of the function f(x)= 4x^3 - 3x^2- 8x + 5 without using a calculator?

    4. If the graph of f: R arrow R crosses the x axis exactly three times, which one of the following rules could not be the rule for f?
    A. y=x(x^2-4)
    B. y=x(x-2)(x+4)(x^2+1)
    C. y=(3-x)(x^4-16)
    D y=(x^2-x-6)(x-4)
    E y=(x^2-x-12)

    5. For the curve of the function with equation y= (2-x)(x-2)(x-5), the subset of R for which the gradient of the graph is positive is best described by?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Feb 2008
    From
    Melbourne
    Posts
    27
    Quote Originally Posted by chaneliman View Post
    1. a. How do u identify the x-intercepts of the graph of y=x(x^2-9a^2) (note: a>0)

    y = (x+0) (x -3a)(x+3a)

    x intercepts are

    x+0 = 0 , x = 0
    x-3a=0 , x= 3a
    x+3a =0 , x= -3a
    Quote Originally Posted by chaneliman View Post
    b. How would u go about sketching the graph of y=
    x(x^2-9a^2)
    power of 3 we start with a positive gradient … unless there was a reflection in the x

    from left to right pass though -3a 1st then loop back 0 then loop back around to 3a …

    Quote Originally Posted by chaneliman View Post
    2. If the equation 2x^2 + 6=ax has exactly one solution, find the value(s) of a, such that this is true.
    2x^2 –ax + 6 = 0

    then take the Discriminate and set it to 0

    b^2 - 4ac = 0

    a^2 - 4(12)= 0

    a^2 = 48

    a= +root48 or –root48


    Quote Originally Posted by chaneliman View Post
    3. Is it possible to find the x intercepts of the function f(x)= 4x^3 - 3x^2- 8x + 5 without using a calculator?
    yes … but you will be here all day … and need to know polynomial division which would take a while to teach.. best thing to do .. solve for y=0
    Quote Originally Posted by chaneliman View Post
    4. If the graph of f: R arrow R crosses the x axis exactly three times, which one of the following rules could not be the rule for f?
    A. y=x(x^2-4)
    B. y=x(x-2)(x+4)(x^2+1)
    C. y=(3-x)(x^4-16)
    D y=(x^2-x-6)(x-4)
    E y=(x^2-x-12)
    E because you can only see power of 2… if you expand it so max amount of crosses is 2 …
    Quote Originally Posted by chaneliman View Post
    5. For the curve of the function with equation y= (2-x)(x-2)(x-5), the subset of R for which the gradient of the graph is positive is best described by?
    ok rearrange y= (-x+2)(x-2)(x-5)

    y= -(x-2)(x-2)(x-5)
    y= -(x-2)^2(x-5)

    Intercepts are … 2 and 5 and 2 is a turning point

    ok because its power of the graph is 3 we start positive but we have reflection in x so we actually start negative… we get to 2 bounce up doesn’t not cross remember then we start to get a positive gradient sadly… it returns to pass though 5 … so we need to differentiate the graph…

    y= (-x+2)(x-2)(x-5) = -x^3 + 9x2 -24x+20

    dy/dx = -3x^2+18x-24

    let that equal to 0 to find stationary points

    -3x^2+18x-24=0

    x^2-6x+8=0

    (x-2)(x-4)=0

    therefore 2 and 4 … and that’s where the graph is positive … 2 < x < 4
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Dec 2007
    Posts
    41
    thanks for doing all that. just one question wot do u mean by "because you can only see power of 2… if you expand it so max amount of crosses is 2 …"
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Feb 2008
    From
    Melbourne
    Posts
    27
    Quote Originally Posted by chaneliman View Post
    thanks for doing all that. just one question wot do u mean by "because you can only see power of 2… if you expand it so max amount of crosses is 2 …"
    yer sorry didnt really proof read ... i ment expand the brackets and look at the highest power.. in this case it was x^2 and power of 2 means it crosses the x axis 2 times...

    to cross three times you needed at least a power of three e.g x^3
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Dec 2007
    Posts
    41
    can someone help me find the equation of the following
    Attached Thumbnails Attached Thumbnails Functions help (test soon)-untitled.bmp  
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,807
    Thanks
    116
    Quote Originally Posted by chaneliman View Post
    can someone help me find the equation of the following
    I have to make a few assumptions:

    1. The function is at least a cubic function.
    2. (2, 2) is a turning point.That means f(2) = 2 and f''(2) = 0
    3. At (2, 2) the graph has a horizontal tangent. That means f'(2) = 0
    4. The graph passes through the origin. That means f(0) = 0

    If so:

    Use the genral equation of a cubic function:

    f(x)=ax^3+bx^2+cx+d

    f'(x)=3ax^2+2bx+cx

    f''(x)=6ax+2b

    From #4 you know d = 0

    Plug in the coordinates (2, 2) into the equations and solve the following system of simultaneous equations for (a, b, c):

    \left| \begin{array}{l}8a + 4b + 2c = 2 \\ 12a + 4b + c = 0 \\ 12·a + 2·b = 0 \end{array} \right. . After a few steps you should get (a, b, c) = \left( \frac14 , -\frac32 , 3 \right)
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by chaneliman View Post
    can someone help me find the equation of the following
    Assuming that the function is a cubic and assuming (2, 2) is a stationary point of inflexion (earboth, I'm sure that's the word you meant rather than turning point), the following model can be immediately used:

    f(x) = a(x - 2)^3 + 2.

    To get a, the assumption that (0, 0) is point on the curve will do the trick:

    0 = a(0 - 2)^3 + 2 \Rightarrow a = \frac{1}{4}.

    Then f(x) = \frac{1}{4} (x - 2)^3 + 2.

    This answer is equivalent to the one found by earboth.

    By the way, the point (4, 4) also appears to lie on the curve - our answers are certainly consistent with that.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Junior Member
    Joined
    Dec 2007
    Posts
    41
    thanks. Could u clear some things up for me?

    1. why does the equation x^2-3x move to the right by 3? (i mean its not a translation)

    2. how would u describe the tanformations of the graph y=3x^3 + 5x + x +1? (like dilations, transations, reflections)

    2. has 4/x been dilated by the y-axis (not the x-axis, i know how to do it from the x axis) by a factor of 4 or 1/4?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by chaneliman View Post
    thanks. Could u clear some things up for me?

    1. why does the equation x^2-3x move to the right by 3? (i mean its not a translation) Mr F says: Move to the right relative to what graph?? The question makes no sense.

    2. how would u describe the tanformations of the graph y=3x^3 + 5x + x +1? (like dilations, transations, reflections)

    Mr F says: Transformations compared to what graph?? Again, the question makes no sense.

    2. has 4/x been dilated by the y-axis (not the x-axis, i know how to do it from the x axis) by a factor of 4 or 1/4?

    Mr F says: Dilation from the y-axis: f(x) --> f(ax) and the dilation factor is 1/a.

    f(x) = 1/x --> f(x/4) = 4/x. a = 1/4 and so the dilation factor is 4.
    ..
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Junior Member
    Joined
    Dec 2007
    Posts
    41
    1. relative to x^2

    2. relative to x^3
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by chaneliman View Post
    1. relative to x^2

    Mr F says: Complete the square: x^2 - 3x = \left( x - \frac{3}{2} \right)^2 - \frac{9}{4}.

    So  x^2 \rightarrow x^2 - 3x = \left( x - \frac{3}{2} \right)^2 - \frac{9}{4}.

    You should be able to easily read off the transformations involved.



    2. relative to x^3

    Mr F says: Well then, bad luck sport. You see, the thing is that y = 3x^3 + 5x^2 + x +1 has turning points (at x = -1 and x = -1/9). But y = x^3 has a stationary point of inflexion. It's therefore impossible to get y = 3x^3 + 5x^2 + x +1 from y = x^3 using any combination of translations, dilations or reflection in coordinate axes. So that's that.
    ..
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Junior Member
    Joined
    Dec 2007
    Posts
    41
    1. The largest domain which the function f(x) = 1/square root x-4 is defined by, is: [4, ~) or (4,~ ) and how did u noe?

    2. For the curve of the function with the equation y=(2-x)(x-2)(x-5), the subset of R for whcih the gradient of the graph is positive is best described by?

    3. The parabola with the equation y=x^2 is translated so that its image has a vertex (a,b). The equation of the image is:

    4. The linear factors of x^2-9x are: I said it was 0, -3, 3 but the books says thats wrong and the answer in x, x-3, x+3

    5. 3 points have coordinates A (1,7), B (7,5), C(0, -2). Find the equation of the line perpendicular to AB, passing through the midpoint odf BC?

    6. Given that f(x+4) = x^3-4, then f(x) is? i said it was (x+4)^3-4 but the book says its (x-4)^3-4 y is that?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. The integral test and decreasing functions...
    Posted in the Calculus Forum
    Replies: 3
    Last Post: February 23rd 2010, 12:53 PM
  2. distributions (test functions)
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: September 5th 2009, 08:49 PM
  3. Monotonic Functions and first derivative test
    Posted in the Calculus Forum
    Replies: 2
    Last Post: December 25th 2008, 03:45 AM
  4. functions...test tomorrow =/
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: October 2nd 2008, 07:21 PM
  5. Quadratic Functions - Test Tomorrow
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: September 15th 2008, 12:36 PM

Search Tags


/mathhelpforum @mathhelpforum