57315-0136147054_02

57315-0136147054_02 - Section 2.1 2.1.1: f (x) = 0 x2 + 0 x...

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Section 2.1 2.1.1: f ( x )=0 · x 2 +0 · x +5,so m ( a · 2 · a 0. In particular, m (2) = 0, so the tangent line has equation y 5=0 · ( x 0); that is, y 5. 2.1.2: f ( x · x 2 +1 · x +0,so m ( a · 2 · a 1. In particular, m (2) = 1, so the tangent line has equation y 2=1 · ( x 2); that is, y = x . 2.1.3: Because f ( x )=1 · x 2 · x + 0, the slope-predictor is m ( a )=2 · 1 · a +0=2 a . Hence the line L tangent to the graph of f at (2 ,f (2)) has slope m (2) = 4. So an equation of L is y f (2) = 4( x 2); that is, y =4 x 4. 2.1.4: Because f ( x )= 2 x 2 · x + 1, the slope-predictor for f is m ( a · ( 2) a +0= 4 a .Thu sthe line L tangent to the graph of f at (2 (2)) has slope m (2) = 8 and equation y f (2) = 8( x 2); that is, y = 8 x +9. 2.1.5: Because f ( x · x 2 +4 x 5, the slope-predictor for f is m ( a · 0 · a +4 = 4. So the line tangent to the graph of f at (2 (2)) has slope 4 and therefore equation y 3=4( x 2); that is, y x 5. 2.1.6: Because f ( x · x 2 3 x +7, the slope-predictor for f is m ( a · 0 · a 3= 3. So the line tangent to the graph of f at (2 (2)) has slope 3 and therefore equation y 1= 3( x 2); that is, y = 3 x +7. 2.1.7: Because f ( x x 2 3 x + 4, the slope-predictor for f is m ( a · 2 · a 3=4 a 3. So the line tangent to the graph of f at (2 (2)) has slope 5 and therefore equation y 6=5( x 2); that is, y =5 x 4. 2.1.8: Because f ( x )=( 1) · x 2 3 x + 5, the slope-predictor for f is m ( a · ( 1) · a 2 a 3. So the line tangent to the graph of f at (2 (2)) has slope 7 and therefore equation y +5= 7( x 2); that is, y = 7 x 2.1.9: Because f ( x x 2 +6 x , the slope-predictor for f is m ( a )=4 a + 6. So the line tangent to the graph of f at (2 (2)) has slope m (2) = 14 and therefore equation y 20 = 14( x 2); that is, y =14 x 8. 2.1.10: Because f ( x 3 x 2 +15 x , the slope-predictor for f is m ( a 6 a + 15. So the line tangent to the graph of f at (2 (2)) has slope m (2) = 3 and therefore equation y 18 = 3( x 2); that is, y =3 x +12. 2.1.11: Because f ( x 1 100 x 2 +2 x , the slope-predictor for f is m ( a 2 100 a + 2. So the line tangent to the graph of f at (2 (2)) has slope m (2) = 1 25 +2= 49 25 and therefore equation y 99 25 = 49 25 ( x 2); that is, 25 y =49 x +1. 48
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2.1.12: Because f ( x )= 9 x 2 12 x , the slope-predictor for f is m ( a 18 a 12. So the line tangent to the graph of f at (2 ,f (2)) has slope m (2) = 48 and therefore equation y +60= 48( x 2); that is, y = 48 x + 36. 2.1.13: Because f ( x )=4 x 2 + 1, the slope-predictor for f is m ( a )=8 a . So the line tangent to the graph of f at (2 (2)) has slope m (2) = 16 and therefore equation y 17 = 16( x 2); that is, y =16 x 15. 2.1.14: Because f ( x )=24 x , the slope-predictor for f is m ( a ) = 24. So the line tangent to the graph of f at (2 (2)) has slope m (2) = 24 and therefore equation y 48 = 24( x 2); that is, y =24 x . 2.1.15: If f ( x x 2 + 10, then the slope-predictor for f is m ( a 2 a . A line tangent to the graph of f will be horizontal when m ( a )=0,thuswhen a = 0. So the tangent line is horizontal at the point (0 , 10) andatnootherpo into fthegrapho f f .
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57315-0136147054_02 - Section 2.1 2.1.1: f (x) = 0 x2 + 0 x...

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