Results 1 to 12 of 12

Math Help - Sketch graph of function f(x) = 2(x-1)^2 - 3

  1. #1
    Member
    Joined
    Jun 2009
    Posts
    86

    Sketch graph of function f(x) = 2(x-1)^2 - 3

    The question states: Determine the domain and range of the function f(x) = 2(x-1)^2 - 3 by sketching its graph.

    I know how to determine D and R, though I forgot how to sketch parabolas on graph. I think I also forgot how to factor too...

    So far I did the following:

    f(x) = 2(x - 1)^2 - 3
    0 = 2(x - 1)^2 - 3
    0 = 2(x - 1)(x - 1) - 3
    0 = 2x^2 - 1x -1x + 2 - 3
    0 = 2x^2 - 2x - 1

    What do I do from here? Is this correct so far, or did I mess up on the first line (with the 2)?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Feb 2009
    From
    stgo
    Posts
    84
     f(x) = 2(x-1)^2 - 3

    At first, note that the coefficient that multiplies to x^2, is postive. With this information, you know that the parabola opens as a smile. Now, where's the vertex? It's on (1,-3) (can you tell us why?). Now, you can sketch the graph of the function. You'll know that the function takes all the values greater or equal to the vertex. Which would be the range?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Jun 2009
    Posts
    86


    "At first, note that the coefficient that multiplies to x^2, is postive. With this information, you know that the parabola opens as a smile."

    Had this already; a positive 2 = minimum value.


    "Now, where's the vertex? It's on (1,-3) (can you tell us why?)."

    This is where I got confused. The -3 is the y value, though I wasn't sure as to how to get the x-value. My guess is that the brackets must = 0? Therefore 1 - 1 = 0?


    Now, you can sketch the graph of the function. You'll know that the function takes all the values greater or equal to the vertex. Which would be the range?

    Think I got it. Because the two x intercepts can be anywhere on the x axis, then the domain = {xER}. And because the parabola opens where y = -3, any y coordinate below -3 is not part of the graph, making the range = {ER|y>=-3}.

    This look correct? (Just need clarification on the x value = 1.)
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Feb 2009
    From
    stgo
    Posts
    84
    Yes, your right with the range!

    The x is not restricted by anything, and the y can't reach values below -3. To solve the confussions you have, read this article

    Parabola - Wikipedia, the free encyclopedia
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Jun 2009
    Posts
    86
    Thx.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Jun 2009
    Posts
    86
    Another question is asking:

    Determine the range if the domain is {-3, -1, 0, 2.5, 6}

    a) f(x) = 4-3(x) b) f(x) = 2x^2 - 2x + 1

    Do I just sub the x values into the equation to solve for the y values?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member
    Joined
    Jun 2009
    From
    Africa
    Posts
    641

    Smile

    Quote Originally Posted by shadow6 View Post
    Another question is asking:

    Determine the range if the domain is {-3, -1, 0, 2.5, 6}

    a) f(x) = 4-3(x) b) f(x) = 2x^2 - 2x + 1

    Do I just sub the x values into the equation to solve for the y values?
    hello
    the range would be, \left \{ {f(-3), f(-1), f(0), f(2.5), f(6)} \right \}.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member
    Joined
    Jun 2009
    Posts
    86
    For which question, a or b? And how do you get the range? Are they they same as the domain or do you have to sub something?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Super Member
    Joined
    Jun 2009
    From
    Africa
    Posts
    641

    Smile

    Quote Originally Posted by shadow6 View Post
    For which question, a or b? And how do you get the range? Are they they same as the domain or do you have to sub something?
    for any function defined on those points.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Member
    Joined
    Jun 2009
    Posts
    86
    Alright thx.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Member
    Joined
    Jun 2009
    Posts
    86
    Another question gives me an equation f(x) = -2(x + 1)^2 + 3

    Then it tells me to determine:

    i) f(1) - f(0)

    ii) 3f(2) - 5

    iii) f(2 - x)


    I started with the first.

    I subbed f(1) - f(0) into f(x) = -2(x + 1)^2 + 3.

    f(x) = -2(x + 1)^2 + 3
    f(1) - f(0) = -2((1 - 0) + 1)^2 + 3
    f(1) - f(0) = -2(1 + 1)^2 + 3
    f(1) - f(0) = -2(2)^2 + 3
    f(1) - f(0) = -2(4) + 3
    f(1) - f(0) = -8 + 3
    f(1) - f(0) = -5

    That look right? Did I sub correctly?
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Super Member
    Joined
    Jun 2009
    From
    Africa
    Posts
    641

    Smile

    Quote Originally Posted by shadow6 View Post
    Another question gives me an equation f(x) = -2(x + 1)^2 + 3

    Then it tells me to determine:

    i) f(1) - f(0)

    ii) 3f(2) - 5

    iii) f(2 - x)


    I started with the first.

    I subbed f(1) - f(0) into f(x) = -2(x + 1)^2 + 3.

    f(x) = -2(x + 1)^2 + 3
    f(1) - f(0) = -2((1 - 0) + 1)^2 + 3
    f(1) - f(0) = -2(1 + 1)^2 + 3
    f(1) - f(0) = -2(2)^2 + 3
    f(1) - f(0) = -2(4) + 3
    f(1) - f(0) = -8 + 3
    f(1) - f(0) = -5

    That look right? Did I sub correctly?
    it doesn't look right.
    f(1)=-5.
    f(0)=1.
    f(1)-f(0)=-5-1=-6.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. dis/continuous function - Sketch Graph
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: March 14th 2011, 01:18 PM
  2. Sketch the graph of each function
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: February 17th 2011, 03:46 PM
  3. Sketch the graph of each function
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: February 17th 2011, 03:11 PM
  4. Sketch the graph of a continuous function
    Posted in the Calculus Forum
    Replies: 2
    Last Post: December 12th 2009, 10:50 AM
  5. Sketch the graph of a function
    Posted in the Pre-Calculus Forum
    Replies: 6
    Last Post: September 30th 2007, 02:45 PM

Search Tags


/mathhelpforum @mathhelpforum