
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 m_{1} and m_{2}.

The lines are parallel if they have the same slope,
that is
m_{1}
= m_{2}

The lines are perpendicular
if
m_{1} = 1/m_{2}

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 pointslope 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 xcoordinate 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

Quadratic Functions in Standard Form
Recall what the function
y = x^{2}
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 = 2x^{2 } 8x + 2

Factor the leading coefficient:
y =
2(x^{2 } 4x + 1)

Calculate
b/2:
(4)/2 = 2

Square the solution above:
2^{2}
= 4

Add and subtract answer from part three inside
parentheses:
y = 2(x^{2 }
4x + 4  4 + 1)

Regroup:
y
= 2[(x^{2 } 4x + 4 )  3]

Factor the inner parentheses using part two as a hint:
y = 2[(x 
2)^{2}  3]

Multiply out the outer constant:
y = 2(x  2)^{2}
 6

Applications:
Newton's law of motion:
Newton discovered that if an object is thrown from an initial height of
s_{0} feet with an initial velocity of v_{0} feet per second
then the position of the function of the object is
s = 16t^{2} + v_{0}t + s_{0
}
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
v_{0}
= 25 and s_{0} = 5
so that
s = 16t^{2} + 25t + 5
When it hits the ground
s = 0
or
0 = 16t^{2}
+ 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
 Factor the leading coefficient:
y =
16(x^{2 } 25/16 x  5/16)
 Calculate
b/2:
(25/16)/2 = 25/32
 Square the solution above:
(25/32)^{2} = 0.61
 Add and subtract answer from part three inside
parentheses:
y =
16(x^{2 } 25/16 x + 0.61  0.61 + 5/16)
 Regroup:
y
= 16[(x^{2 } 25/16 x + 0.61 )  0.61]

Factor the inner parentheses using part two as a hint:
y = 16[(x 
25/32)^{2}  0.61]

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 < 16t^{2} + 25t + 5
or
0
< 16t^{2} + 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.