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Finding the equation of a parabola with 1 point

  1. Jul 1, 2015 #1
    1. The problem statement, all variables and given/known data


    The sketch on the previous page shows the graph of a function f, which is a parabola with vertex R, and the graph of a function g, which is a straight line defined by g(x)=-(1/2)x. The graphs of f and g intersect at P and O(the origin). The function f is defined by:
    f(x)=ax2+bx+c
    Df=ℝ, Rf=[-2,∞) and b/(2a)=2.


    a)Find the coordinates of R.

    2. Relevant equations
    g(x)=-(1/2)x
    f
    (x)=ax2+bx+c
    Df=ℝ, Rf=[-2,∞) and b/(2a)=2.

    3. The attempt at a solution
    if b/(2a)=2 then my x coordinate of the vertex = 2 correct? but on the graph it looks more like -2
    Is this due to the missing (-)b

    I am stumped on this question, I am sure It is quite obvious but I would appreciate a point in the right direction.
    Thank you,
    Jaco
     

    Attached Files:

    Last edited: Jul 1, 2015
  2. jcsd
  3. Jul 1, 2015 #2

    RUber

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    If b/(2a) = 2, then b = 4a, so you can rewrite f(x) = ax^2 + 4ax + c. This has a minimum where 2a x +4a = 0, or where x=-2. So your original insight was right, but you might have tried to jump a logical step and missed where the negative belongs.

    How can you tell what the y coordinate might be?
     
  4. Jul 1, 2015 #3

    RUber

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    Hint: f(0) = 0.
     
  5. Jul 1, 2015 #4
    ##f(x) = ax^2 + bx + c = a(x-h)^2 + k##

    Given: ##\frac{b}{2a} = 2##

    The range of the function is from -2 to ##\infty##. What does this tell you about the y-coordinate of R?

    http://mathbitsnotebook.com/Algebra1/Quadratics/QDVertexForm.html

    Not only should you be able to find the coordinates of R. You should be able to find the equation for the parabola (and thus find any point on the parabola).
     
  6. Jul 3, 2015 #5

    ehild

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    That is the picture:

    lineparab.JPG
     
  7. Jul 6, 2015 #6
    Rf={-2,∞)
    so y=-2
    so my vertex is (-2,-2) (Coordinates of R)
    f(x)=1/2(x+2)^2-2
     
    Last edited: Jul 6, 2015
  8. Jul 6, 2015 #7

    SammyS

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    Yes, that's correct
     
  9. Jul 6, 2015 #8
    Find:
    a, b and c
    y=1/2(x+2)^2-2
    y=1/2(x+2)(x+2)-2
    y=1/2(x^2+2x+2x+4)-2
    y=1/2x^2+2x+0
    so a=1/2, b=2 and c=0
     
  10. Jul 6, 2015 #9

    SammyS

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    Also is correct.
     
  11. Jul 6, 2015 #10
    Thank you Sammy,
     
  12. Jul 6, 2015 #11
    Find the equation of a line that is parallel to the straight line and passes through R.

    y=-1/2x+0 point R(-2,-2)
    y=mx+b
    -2=-1/2x+b
    -2=-1/2(-2)+b
    -2=1+b
    b=-3

    y=-1/2x-3
     
  13. Jul 6, 2015 #12
    Find the coordinates of P
    (1/2)x^2+2x+0=-(1/2)x+0
    (1/2)x^2+2x+(1/2)x=0
    (1/2)x^2+2&(1/2)x=0
    (x^2)/2+(5x)/2=0
    x^2+5x=0
    x(x+5)=0

    x=0 and x=-5

    so (-5,?)

    y=-1/2(x)
    y=-1/2(-5)
    y=2.5

    so P (-5,2.5)
     
    Last edited: Jul 6, 2015
  14. Jul 6, 2015 #13

    Mark44

    Staff: Mentor

    Your answer below is correct. I've added some comments about how you could work more efficiently.
    Instead of dragging those 1/2 factors along, just multiply both sides of the equation by 2, which results in
    ##x^2 + 4x + x = 0##, or ##x^2 + 5x = 0##.

     
  15. Jul 6, 2015 #14
    Thank you Mark,
     
  16. Jul 6, 2015 #15
    Calculate the distance between P and Q
    P(-5,2.5) Q(?,0)
    Find Q

    x intercepts:
    y=(1/2)x^2+2x+0
    ((1/2)x+2)(x+0)
    (1/2)x=-2 and x=0
    x=-4 and x=0

    Q(-4,0)

    d=√{(x2-x1)2+(y2-y1)2}
    d=√{(-4-(-5))2+(0-2.5)2}
    d=√{(-4+5))2+(0-2.5)2}
    d=√{12+(-2.5)2}
    d=√{1+6.25}
    d=√{7.25}

    The distance between P and Q is √{7.25}
     
    Last edited: Jul 6, 2015
  17. Jul 6, 2015 #16
    I do not understand the following question, some help please.

    Calculate the maximum vertical distance between corresponding points on the graphs of f and g on the interval[xP,0], where xP denotes the x-coordinate of P.
     
  18. Jul 6, 2015 #17

    Mark44

    Staff: Mentor

    The vertical distance between the line and the parabola is ##h = -\frac x 2 - \frac{x^2}{2} - 2x##. This distance function turns out also to be a quadratic, so you can find its maximum value by techniques you already know.
     
  19. Jul 9, 2015 #18
    hello Jaco and Sammy i have a question i can see on the equation you put 1/2 (x+2)^2-2 i am tryna find out where the 1/2 came from looking from the g=-1/2x does it mean when its parallel the -1/2 becomes positive 1/2
     
  20. Jul 9, 2015 #19
    Hi Toni,
    ƒ(g) is a line,
    f(x)=1/2(x+2)^2-2 is the parabola.

    so no.

    I used:
    ƒ(x)=a(x-h)^2+k to find a, 1/2(x+2)^2-2
     
  21. Jul 9, 2015 #20
    thanks i got it
     
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