Results 1 to 3 of 3

Math Help - Functions and finding x-intercepts and vertex.

  1. #1
    Newbie
    Joined
    Mar 2010
    Posts
    1

    Question Functions and finding x-intercepts and vertex.

    Hi, I'm new to Math Help Forum and was wondering if you could help me with functions and finding the x-intercepts and the coordinates of the vertex.

    I'm having a hard time with these questions:

    y=x^2+7x

    and

    y= 15x^2+16-7

    Can you please explain step by step. I use the quadratic formula but my anwser seems be be wrong.

    thanks for taking time to read this, hopefully you can help.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,829
    Thanks
    123
    Quote Originally Posted by kimmybimmy View Post
    Hi, I'm new to Math Help Forum and was wondering if you could help me with functions and finding the x-intercepts <<<<<< y = 0
    and the coordinates of the vertex. <<<<<<<< complete the square

    I'm having a hard time with these questions:

    y=x^2+7x

    and

    y= 15x^2+16-7

    Can you please explain step by step. I use the quadratic formula but my anwser seems be be wrong.

    thanks for taking time to read this, hopefully you can help.
    I'll take the 2nd example, then the 1st example is left for you:

    1. I assume that you mean:

    y = 15x^2+16x-7

    2. x-intercept: y = 0

    0=15x^2+16x-7 Use the quadratic formula

    x = \frac{-16 \pm \sqrt{(-16)^2-4 \cdot 15 \cdot (-7)}}{2 \cdot 15} Simplify!

    I've got x = \frac13~\vee~x=-\frac75

    3. Coordinates of the vertex:

    y=15x^2+16x-7 = 15\left(x^2+\frac{16}{15}x\bold{\color{red}+\left(  \frac8{15}\right)^2-\left(\frac8{15}\right)^2}\right) - 7

    y=15x^2+16x-7 = 15\left(\left(x+\bold{\color{red}\left(\frac8{15}\  right)}\right)^2 \bold{\color{red}-\left(\frac8{15}\right)^2}\right) - 7

    y=15x^2+16x-7 = 15\left(x+\bold{\color{red}\left(\frac8{15}\right)  }\right)^2 \bold{\color{red}-\left(\frac{64}{15}\right)} - 7

    y=15x^2+16x-7 = 15\left(x+\left(\frac8{15}\right)\right)^2 -\frac{169}{15}

    Therefore the vertex is at V\left(-\frac8{15}~,~-\frac{169}{15}\right)

    4. And now it's your turn!
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,319
    Thanks
    1294
    Since you going to have to complete the square to find the vertex anyway, I would prefer using that to find the y- intercepts rather than the quadratic formula.

    y= x^2+ 7x= 0

    Half of 7 is 7/2 and the square of that is 49/4:

    y= x^2+ 7x+ \frac{49}{4}- \frac{49}{4}= 0

    y= (x+ \frac{7}{2})^2- \frac{49}{4}= 0

    At this point you can argue that (x+ \frac{7}{2})^2 is never negative so y is always -\frac{49}{4} "plus something". That tells you than the lowest value for y is -49/4 which occurs when x+ \frac{7}{2}= 0 or  x= -\frac{7}{2}.

    The vertex is at \left(-\frac{7}{2}, -\frac{49}{4}\right)

    Now, going back to (x+ \frac{7}{2})^2- \frac{49}{4}= 0,
    adding \frac{49}{4} to each side gives
    (x+ \frac{7}{2})^2= \frac{49}{4}
    and so

    x+ \frac{7}{2}= \pm\frac{7}{2}

    so that x= -\frac{7}{2}+ \frac{7}{2}= 0
    or x= -\frac{7}{2}- \frac{7}{2}= -7.

    Of course, for this problem, the simplest way to do it is to factor:
    x^2+ 7x= x(x+ 7)= 0 so either x= 0 or x+ 7= 0 and x= -7.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: November 3rd 2009, 09:17 AM
  2. Finding vertex
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: January 20th 2009, 07:55 PM
  3. finding the vertex of quadratic functions
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: October 18th 2008, 09:58 AM
  4. Replies: 1
    Last Post: May 20th 2008, 09:56 PM
  5. finding the vertex
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: July 17th 2007, 09:41 PM

Search Tags


/mathhelpforum @mathhelpforum