Results 1 to 8 of 8

Math Help - Quadratic Functions

  1. #1
    Newbie
    Joined
    Oct 2008
    Posts
    21

    Quadratic Functions

    Another function problem...

    Suppose I have the function f(x)=x^2-1 how would I find the intercepts?

    The vertex is easy to find by completing the square. y=a(x-h)^2+k where a= h=0 and k=-1 giving the location of the vertex at (0,-1) Could someone explain how I get them? I keep getting the wrong answer--

    Bam bah Lam--
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    y=x^{2}-1

    The vertex is easier found by using x=\frac{-b}{2a}

    In this case, a=1 and b=0, c=-1.

    Did you notice this is a difference of two squares.

    y=x^{2}-1=(x+1)(x-1)

    If you want the y-intercept, set x = 0 and solve. If you want the x-interceot, set y=0 and solve.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    A riddle wrapped in an enigma
    masters's Avatar
    Joined
    Jan 2008
    From
    Big Stone Gap, Virginia
    Posts
    2,551
    Thanks
    11
    Awards
    1
    Quote Originally Posted by WhoaBlackBetty View Post
    Another function problem...

    Suppose I have the function f(x)=x^2-1 how would I find the intercepts?

    The vertex is easy to find by completing the square. y=a(x-h)^2+k where a= h=0 and k=-1 giving the location of the vertex at (0,-1) Could someone explain how I get them? I keep getting the wrong answer--

    Bam bah Lam--
    To find the y-intercept, find f(0).

    x^2-1=0
    f(0)=0^2-1
    f(0)=-1. This is your y-intercept.

    To find the x-intercepts (zeros of the function), set f(x)=0 and solve for x.

    x^2-1=0
    (x-1)(x+2)=0
    x=1 \ \ or \ \ x=-1 These are your x-intercepts.

    Too fast for me Galactus!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Oct 2008
    Posts
    21
    Ok, that makes more since. This is confusing for some reason to me. Harder than just solving the quadratic. So, if I was going to graph it how would I go about that? I know my first point the vertex would be at (0,-1) So, I would have a point there, but would I just use the x-intercepts combined with plugging in a Y value for the other two points? Or something different? Thanks...
    Follow Math Help Forum on Facebook and Google+

  5. #5
    A riddle wrapped in an enigma
    masters's Avatar
    Joined
    Jan 2008
    From
    Big Stone Gap, Virginia
    Posts
    2,551
    Thanks
    11
    Awards
    1
    Quote Originally Posted by WhoaBlackBetty View Post
    Ok, that makes more since. This is confusing for some reason to me. Harder than just solving the quadratic. So, if I was going to graph it how would I go about that? I know my first point the vertex would be at (0,-1) So, I would have a point there, but would I just use the x-intercepts combined with plugging in a Y value for the other two points? Or something different? Thanks...
    Set up a table using arbitrary x values. Then find f(x) to complete the ordered pair. Then plot them.

    f(x)=x^2-1

    x=0, f(0)=-1, Plot (0, -1)
    x=1, f(1)= 0, Plot (1, 0)
    x=-1, f(-1)=0, Plot (-1, 0)
    .
    .
    .
    etc.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Oct 2008
    Posts
    21
    Thank you. I think I understand it now!
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Oct 2008
    Posts
    21
    One more, I am doing something wrong. I know what the values are supposed to be like (2,4) (3,1) (4,0) (5,1) (6,4)...etc. How do I plug the values in to get those numbers? It must be something simple I am missing. Here is the problem f(x)=(x-4)^2

    This is taking me longer than usual to wrap my head around... So, I know this is a horizontal shift with the vertex at (4,0) The part I am not getting is plugging it into the equation. Take this for example--if

    f(x)= (x-4)^2 with x= to 1

    f(1)= (1-4)(1-4) Or, am I mathing these out wrong? Doing this way you wind up with 1 and 9 unless you take the square which is 1,3 ....
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    Oct 2008
    Posts
    21
    Anyone...
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. quadratic functions
    Posted in the Algebra Forum
    Replies: 4
    Last Post: May 2nd 2010, 10:04 PM
  2. Quadratic Functions
    Posted in the Algebra Forum
    Replies: 1
    Last Post: January 15th 2010, 03:10 AM
  3. quadratic functions
    Posted in the Algebra Forum
    Replies: 4
    Last Post: November 23rd 2009, 01:35 PM
  4. Quadratic Functions
    Posted in the Algebra Forum
    Replies: 16
    Last Post: July 27th 2009, 10:11 AM
  5. Quadratic functions
    Posted in the Algebra Forum
    Replies: 3
    Last Post: November 14th 2007, 05:11 PM

Search Tags


/mathhelpforum @mathhelpforum