Results 1 to 10 of 10

Math Help - quadratic function and vertex help!

  1. #1
    Member
    Joined
    Mar 2012
    From
    nyc
    Posts
    120
    Thanks
    1

    quadratic function and vertex help!

    ok this is a very confusing topic. im doing the review for my midterm and encountered a problem, heres the question:

    g(x)=x^2-2x+5 professors answer =vertex is (1,4) and x-intercept is none and y--intercept is (0,5)

    the only thing i got right was the y-intercept. for the vertex im getting (1,6), and why is there not an x-intercept?

    id appreciate any help? thank u!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member beebe's Avatar
    Joined
    Aug 2011
    Posts
    54

    Re: quadratic function and vertex help!

    You got the x part of the vertex correct. To find the y coordinate, plug x=1 back in. What's g(1)?

    You find the x-intercept by setting g(x)=0 and solving. Can you solve it? What happens when you use the quadratic formula to find the x-intercepts. Also, you may notice that the coefficient on the x^2 is positive, which means the parabola opens upwards, and the vertex is above the x-axis (4 is a positive number). Since the vertex is the lowest point of the parabola and it only goes up from there, it can never cross the x-axis.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Nov 2011
    From
    Crna Gora
    Posts
    420
    Thanks
    64

    Re: quadratic function and vertex help!

    Quote Originally Posted by noork85 View Post
    ok this is a very confusing topic. im doing the review for my midterm and encountered a problem, heres the question:

    g(x)=x^2-2x+5 professors answer =vertex is (1,4) and x-intercept is none and y--intercept is (0,5)

    the only thing i got right was the y-intercept. for the vertex im getting (1,6), and why is there not an x-intercept?

    id appreciate any help? thank u!
    g(x)=ax^2+bx+c

    hence :

    a=1 , ~ b=-2, ~c=5

    therefore :

    D=b^2-4ac=-16

    for vertex we have :

    V\left(\frac{-b}{2a},\frac{-D}{4a}\right) \Rightarrow V(1,4)

    Since D < 0 graph of the function and x-axis have no common points .
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Mar 2012
    From
    nyc
    Posts
    120
    Thanks
    1

    Re: quadratic function and vertex help!

    can i use this formula for all that are in the format of g(x)=ax^2+bx+c?

    Quote Originally Posted by princeps View Post
    g(x)=ax^2+bx+c

    hence :

    a=1 , ~ b=-2, ~c=5

    therefore :

    D=b^2-4ac=-16

    for vertex we have :

    V\left(\frac{-b}{2a},\frac{-D}{4a}\right) \Rightarrow V(1,4)

    Since D < 0 graph of the function and x-axis have no common points .
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Mar 2012
    From
    nyc
    Posts
    120
    Thanks
    1

    Re: quadratic function and vertex help!

    and im still confused about the x-intercept. i know im slow....but ...please help!

    and whats the D stand for? i know is vertex?

    still lost, where does the -16 come from? can u break down step by tiny step?
    Last edited by noork85; March 21st 2012 at 10:15 PM.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member beebe's Avatar
    Joined
    Aug 2011
    Posts
    54

    Re: quadratic function and vertex help!

    D is the discriminant. If you haven't learned it yet, when D>0, then there are two roots. When D=0, there is one root. When D<0, then there are no roots.

    Notice that the discriminant is the part inside of the square root for the quadratic formula. x=\frac{-b \pm \sqrt {b^2-4ac}}{2a} As princeps pointed out, b^2-4ac is a negative number. You can't take a square root of a negative number and get a real number for a result. In other words, you can't square any real number and get -16.

    Step by step:
    D=b^2-4ac
    You know what a, b, and c are, so just substitute and do the arithmetic.
    D=(-2)^2-4*1*5=4-20=-16

    Alternately, if you had gone straight to the quadratic formula to find the x-intercepts, you would get:
    x=\frac{-b \pm \sqrt {b^2-4ac}}{2a}
    x=\frac{-(-2) \pm \sqrt {(-2)^2-4*1*5}}{2*1}
    x=\frac{2 \pm \sqrt{-16}}{2}
    The quadratic formula will always give you the x-intercepts for a quadratic function (if they exist), but you can't find the square root of -16, so there are no real number x-intercepts.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Mar 2012
    From
    nyc
    Posts
    120
    Thanks
    1

    Re: quadratic function and vertex help!

    oh thank u so much! that really cleared it up. i didnt learn anything about discriminitives. but yes, i shouldve just gone straight to the quadratic formula to solve it.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member
    Joined
    Mar 2012
    From
    nyc
    Posts
    120
    Thanks
    1

    Re: quadratic function and vertex help!

    can u also please explain the vertex part.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Junior Member beebe's Avatar
    Joined
    Aug 2011
    Posts
    54

    Re: quadratic function and vertex help!

    It might be helpful if you showed how you got to your answer.

    You found the x value for the vertex already. x=1 is correct. Let's split the denominator of the quadratic formula so that it's x=\frac{-b}{2a}\pm\frac{\sqrt{b^2-4ac}}{2a}. Notice that you add or subtract \frac{\sqrt{b^2-4ac}}{2a} from \frac{-b}{2a}. Also remember that the vertex of a parabola is on it's axis of symmetry. So if you add or subtract the same amount from a number to get the x-intercepts, then the axis of symmetry is at x=\frac{-b}{2a}.

    Plug in the numbers and do the arithmetic and you get x=1, which you already knew. At this point, noork85 went a seperate route to find the y value, which I'm not familiar with, so here's the way that I do it, which seems more intuitive:

    Where is the parabola at when x=1? It's at g(1). All you have to do is go back to your original function and plug 1 in for x and do the arithmetic.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Member
    Joined
    Mar 2012
    From
    nyc
    Posts
    120
    Thanks
    1

    Re: quadratic function and vertex help!

    thank u so very much! very detailed and helpful!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: August 25th 2011, 10:08 AM
  2. Replies: 5
    Last Post: April 16th 2011, 04:11 PM
  3. Replies: 5
    Last Post: March 12th 2010, 01:47 PM
  4. Identifying the vertex of a Quadratic Function
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: October 10th 2009, 10:53 AM
  5. Replies: 5
    Last Post: August 8th 2009, 06:04 PM

Search Tags


/mathhelpforum @mathhelpforum