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Quadratic Equation Straight Line

  1. Oct 24, 2004 #1

    Please Can You Help Me To Understand How To Find The Vertex Of A Quadratic Equation (x,y Coordinates) Just By Looking At The Equation ?

    Eg What Is The Vertex Of Y=3x^2+4x+7 ?


    Also, How Do I Transform A Straight Line Graph Left Or Right On The x Axis Without Moving The Line Up/down Along The Y ?


  2. jcsd
  3. Oct 24, 2004 #2


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    Complete the square.

    If y= 3x2+4x+ 7, rewrite it as y= 3(x2+ (4/3)x)+ 7
    A "perfect square" is of the form (x+ a)2= x2+ ax+ a2. Compare that to x2+ (4/3)x. The coefficient of x is 4/3. 2a= 4/3 if a= 2/3 and, in that case, a2= 4/9. If we add 4/9 then x2+ (4/3)x+ 4/9= (x+ 2/3)2, a perfect square.

    Of course, we can't JUST add 4/9, that would change the value. We can, however both add and subtract 4/9:

    y= 3(x2+ (4/3)x+ 4/9- 4/9)+ 7 and now take the "-4/9" out of the parentheses:
    = 3(x2+ (4/3)x+ 4/9)- 4/3+ 7
    = 3(x+ 2/3)2+ 17/3

    If x= -2/3, then x+ 2/3= 0 so y= 17/3. If x is ANY OTHER number, x+ 2/3 is NOT 0 so the square is positive and y is 17/3 PLUS some number: larger than 17/3.

    The lowest point on the parabola, the vertex, is (-2/3, 17/3).

    To "transform a straight line left or right on the x-axis", in other words, translate left or right, ADD OR SUBTRACT SOMETHING FROM x.

    If the equation of the line is y= 5x+ 3, then y= 5(x- 2)+ 3 = 5x- 10+ 3= 5x- 7 has graph a line parallel to y= 5x+ 3 but shifted right 2 places: y= 3 in 5x+ 3 when x= 0 but in y= 5x- 7 when x= 2.

    The graph of any function, y= f(x), is shifted to the right a places is x is replaced by x-a: y= f(x-a).
    Last edited by a moderator: Oct 24, 2004
  4. Oct 24, 2004 #3

    thanx for the help.

    please can you go over finding the vertex..

    I thought the 17/3 bit is simply the y intercept ?

    Also for a straight line graph, is that to say you cant move the line left or right withought affecting the y intercept ?

  5. Oct 24, 2004 #4


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    Here's a calculus-based approach:

    y= 3x2+4x+ 7
    has its extremum when dy/dx=0

    so, 0=(dy/dx)at extremum=6xat extremum+4

    thus, xat extremum=-2/3
    and yat extremum=3(-2/3)2+4(-2/3)+7

    Here's a variant, inspired by kinematics:


    0=v at the maximum height. So, tat extremum=-v0/a.

    Thus, think of the original equation as
    y=(1/2) 6x2+4x+ 7
    so, xat extremum=-(4)/(6).
  6. Oct 24, 2004 #5

    At the minima and maxima points of the curve, dy/dx=0 => 2ax+b=0 => x=-b/2a

    This doesn't work all the time though.
  7. Oct 24, 2004 #6


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    Ah, yes, of course.
    The x-coordinate of the vertex of y=ax2+bx+c is -b/2a,
    which can be interpreted as the average of the roots of "0=ax2+bx+c".
  8. Oct 24, 2004 #7


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    No, the orginal equation was y= 3x2+ 4x+ 7= 0. The y intercept is, by definition, the y coordinate of the point where the graph crosses the y-axis. Since every point on the y-axis has x-coordinate 0, the y intercept is
    y= 3(0)2+ 4(0)+ 7= 7.

    That's exactly what it says. The line y= x+ 1 has y intercept 1. If fact, since it has slope 1, the distance from the x intercept to the origin is exactly the same as the distance from the origin to the y intercept: 1. If I move the line parallel to itself, say to the left 2 spaces, it still has slope 1 (that depends only on the angle) so distance from the x intercept to the origin must still be the same as the distance from the origin to the y intercept.
    Moving to the left 2 spaces the x intercept is now (-3,0) so the y intercept is (0,3).
    Translating a line to the left or right keeps the slope the same, not the intercepts.
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