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# Alg2 Final Keynote

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### Alg2 Final Keynote

1. 1. This be my Algebra 2 Final Keynote. Read and be afraid, no?Tuesday, June 7, 2011
2. 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. 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. 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, 250Tuesday, June 7, 2011
5. 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=cSecond point: Third Point: 9=a(10)^2 + b(10) + 6 15=a(20)^2+b(20)+63=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+6Tuesday, June 7, 2011
6. 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. 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=.000857142857143Tuesday, June 7, 2011
8. 8. Equation: y=(.000857142857143)x^2+(.251428571428571)x+31 Prediction: y=(.000857142857143)(210)^2+ x=210 (.251428571428571)(210)+31 y=37.8+52.79999999999991+31 y=121.6 Prediction: 121.6 (x104)Tuesday, June 7, 2011
9. 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)xTuesday, June 7, 2011
10. 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. 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. 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=abxthree different lines, three different predictions.Tuesday, June 7, 2011
13. 13. Thanks for watching!Tuesday, June 7, 2011