Equations of Lines and Quadratic Functions

I.  Return Midterm I

II.  Homework

IV.  Parallel Lines and Perpendicular Lines

Two line are parallel if they have the same slope and two lines are perpendicular if

m1 = -1/m2  

where m1 and m2 are the slopes of the first and second lines respectively. 

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

IV.  Quadratic Functions in Standard Form

Recall what the function y = x2 looks like.  It is a parabola with vertex at the origin.  We will 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:

y = a(x - h)2 + k  

Where 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:

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

III.  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?  For what times will the ball be higher than 7 feet?

We will demonstrate the solution on a graphing calculator.