1. This be my Algebra 2 Final Keynote.
Read and be afraid, no?
Tuesday, June 7, 2011
2. So you have a line..
You’re given two points.You know the
line is linear (doesn’t bend).
Ex: (1, 2) First of all, the equation for a line is
(2,4) y=mx+b
Slope y-intercept
Plug your two points in a rise/run (or y2-y1/x2-x1)
equation to find slope.
Ex: y2-y1
x2-x1 = 4-2 2 For this example,
2-1 = 1 slope = m = 2
Plug in the slope and one x/y y=mx+b b=0
coordinate pair to find the y-
2=2(1)+b y=2x+0
intercept.
Tuesday, June 7, 2011
3. The first number
plotted is the
population in 1850. It
is the starting point
for this line, so it
touches the y-axis.
ex: (0, 313)
The next date is
1900. That’s 50 years
after 1850.
ex: (50, 513.6)
And so on.
Ex: y=4.1258x+313 (x10^4)*
*(Everything is 1,000 times larger (10^4))
Tuesday, June 7, 2011
4. Population
chart
Pair of coordinates.
Scatter-plot graph.
Finding the slope.
y-y1=m(x-x1)
form
Equation: y=41258.333x+3130000
Prediction: 11, 794, 250
Tuesday, June 7, 2011
5. So you have a quadratic line..
ex: (0,6) You’re given three points.
(10,9) The formula is y=ax^2+bx+c
(20,15)
We plug in the first point first, because of the x being 0.
(6)=a(0)^2 + b(0) + c
6=c
Second point: Third Point:
9=a(10)^2 + b(10) + 6 15=a(20)^2+b(20)+6
3=100a+10b 9=400a^2+20b
b=(3/10)-10a 9=400a+20([3/10]-10a)
3=100(.015)+10b a=.015
b=.15 y=.015x^2+.15x+6
Tuesday, June 7, 2011
6. The first number
plotted is the
population in 1850.
Plugging in 0 for x
makes y=6. So this
line touches the y-
axis at 6.
ex: (0,6)
Example quadratic line is The next date is
y = 0.015x^2 + 0.15x + 6 1900. That’s 50 years
after 1850.
ex: (50, 51)
..And so on.
Tuesday, June 7, 2011
7. From three points to an equation...
(0,31) & (140,83) & (150,88) y=ax^2+bx+c
Point One: Point Two: Point Three:
31=a(0)^2+b(0)+c 83=a(140)^2 88=a(150)^2+b(150)+c
c=31 +b(140)+31 57=22500a+150b
52=19600a 150b=-22500a+57
+140b b=-150a+(57/150)
b=-150(.000857142857143) 52=19600a+140
+(57/150) (-150a+[57/150])
b=.251428571428571 52=19600a-2100a+53.2
-1.2=-1400a
a=.000857142857143
Tuesday, June 7, 2011
9. So you have an exponential curves line..
Ex: You’re given two points.
(6,7)
(8,10) The formula is y=abx
Plug both points in and then substitute.
(7)=ab(6) (10)=ab(8)
a=(7/b6)
10=(7/b6)(b8)
10=7b 2
Plug b in to find a. b=1.195
a=(7/[1.195]6)
a=3.193 ..and you have your equation.
y=(3.193)(1.195)x
Tuesday, June 7, 2011
10. The first number
plotted is the
population in 1850.
By plugging in 0 for
x, b becomes 1, so
y=a.
ex: (0, 32.166)
Example exponential curves line is
y=(32.166214450324878) The next date is
(1.007725795242675)x 1900. That’s 50 years
after 1850.
ex: (50, 47.263)
..and so on.
Tuesday, June 7, 2011
11. From two points to an equation..
(110, 75) & (120, 81) y=abx
Point One: Point One:
75=ab 110 81=ab 120
a=(75/b110) a=(81/b120)
75 81 75=a(1.007725795242675)110
b 110 = b120
75=2.331638997054641a
81=75b 10
a=32.166214450324878
b=1.007725795242675
Equation: y=(32.166214450324878)(1.007725795242675)x
Prediction: (210, 161.919374795460997)
Tuesday, June 7, 2011
12. Why is each prediction different?
The first
equation is linear,
y=mx+b.
The second
equation is
quadratic,
y=ax2+bx+c
The third
equation is
exponential,
Three different formulas, y=abx
three different lines, three
different predictions.
Tuesday, June 7, 2011
