Equations of Lines and Quadratic Functions

  1. Parallel Lines and Perpendicular Lines

    We define parallel and perpendicular lines as follows:

    Definition of Parallel and Perpendicular Lines

    Let two lines be given with slopes m1 and m2.  

    1. The lines are parallel if they have the same slope, that is

                m1  =  m2  

    2. The lines are perpendicular if

                m1 = -1/m2  



    Example:      

    Find the equation of the line that passes through the point (1,4) and is perpendicular to the line 

            3x - 2y = 5

    Solution:  

    First notice that to find the equation of a line we need a point and a slope.  We have the point, namely (1,4).  To get the slope we solve for y in the other line:

            2y = 3x - 5

            y = 3/2 x - 5/2

    Since the slope of the perpendicular line is 3/2, the slope of our line is -2/3.  Now we use the point-slope form:

            y - 4 = 3/2 (x - 1)

            y = 3/2 x +5/2.


    Recall     that a horizontal line is of the form

            y = constant

    and a vertical line is of the form

            x = constant

     

    Example

    Find the equation of the line through the point (5,4) that is vertical


    Solution

    We know that the form of the equation is:

            x = constant

    Since it passes through (5,4) which as x-coordinate 5 that constant must be 5.  We conclude that the equation of the line is 

            x = 5

    To interactively investigate equations of lines go to 

    http://mathcsjava.emporia.edu/~greenlar/ParPerp/ParPerp.html

  2. Quadratic Functions in Standard Form

    Recall what the function 

            y = x2  

    looks like.  It is a parabola with vertex at the origin.  We can use shifting techniques to graph

    y = 3(x - 2)2 + 4

    We see that the vertex is shifted to the right by 2 units and up 4 units.  There is vertical stretching by a factor of three.  In general, we say that a quadratic is in standard form if it looks like:

        Standard Form of a Quadratic

                     y = a(x - h)2 + k  


    Here the h represents the horizontal shift, the k represents the vertical shift, and the a represents the stretching factor.  If a is negative the parabola is concave down (looks like a frown).

    To put a quadratic in standard form, we complete the square:

    Example:

    Put the following in standard form

            y = 2x2 - 8x + 2

    1. Factor the leading coefficient:

              y =  2(x2 - 4x + 1)
    2. Calculate 

              -b/2:  -(-4)/2 = 2
    3. Square the solution above: 

               22  = 4
    4. Add and subtract answer from part three inside parentheses:              

              y = 2(x2 - 4x + 4 - 4 + 1) 
    5. Regroup: 

              y = 2[(x2 - 4x + 4 ) - 3]
    6. Factor the inner parentheses using part two as a hint:                       

              y =  2[(x - 2)2 - 3]
    7. Multiply out the outer constant:   

              y =  2(x - 2)2 - 6


  3. Applications:  

    Newton's law of motion:

    Newton discovered that if an object is thrown from an initial height of s0 feet with an initial velocity of v0 feet per second then the position of the function of the object is

            s = -16t2 + v0t + s0

    Example:

    If you throw a basketball from five feet with an initial velocity of 25 feet per second, how long will it take to hit the ground?  How high will the ball go and when is this maximum height attained?  For what times will the ball be higher than 7 feet?


    Solution

    We have that 

            v0 = 25    and    s0 = 5

    so that 

            s = -16t2 + 25t + 5

    When it hits the ground 

            s = 0

    or 

            0 = -16t2 + 25t + 5

    We get the two solutions 

            t = 1.74 and t = -1.74

    Notice that a negative time means is not the solution, hence the ball hits the ground after 1.74 seconds.

    To find out how high the ball will go, we are after the s coordinate of the vertex.  We put the equation into standard form

    1. Factor the leading coefficient:

              y =  -16(x2 - 25/16 x - 5/16)
    2. Calculate 

              -b/2:  -(25/16)/2 = -25/32
    3. Square the solution above: 

               (-25/32)2 = 0.61
    4. Add and subtract answer from part three inside parentheses:              

              y = -16(x2 - 25/16 x + 0.61 - 0.61 + 5/16) 
    5. Regroup: 

              y = -16[(x2 - 25/16 x + 0.61 ) - 0.61]
    6. Factor the inner parentheses using part two as a hint:                       

              y =  -16[(x - 25/32)2 - 0.61]
    7. Multiply out the outer constant:   

              y =  -16(x - 25/32)2 + 9.77

    Now we can say that the ball will reach a height of 9.77 feet 25/32 seconds after it was released.

    To find out when the ball will be higher than 7 feet we solve

            7 < -16t2 + 25t + 5

    or

            0 < -16t2 + 25t - 2

    Using the quadratic formula we find the roots at 

            t = 0.085 and t = 1.48

    So that the ball will be higher than 7 feet between 0.085 seconds and 1.48 seconds.

Back to Math 103A Home Page

Back to the Math Department Home

e-mail Questions and Suggestions