# 57315-0136147054_02 - Section 2.1 2.1.1 f(x = 0 x2 0 x 5 so...

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Section 2.1 2.1.1: f ( x ) = 0 · x 2 + 0 · x + 5, so m ( a ) = 0 · 2 · a + 0 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 ) = 0 · x 2 + 1 · x + 0, so m ( a ) = 0 · 2 · a + 1 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 + 0 · 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 + 0 · x + 1, the slope-predictor for f is m ( a ) = 2 · ( 2) a + 0 = 4 a . Thus the line L tangent to the graph of f at (2 , f (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 ) = 0 · x 2 +4 x 5, the slope-predictor for f is m ( a ) = 2 · 0 · a +4 = 4. So the line tangent to the graph of f at (2 , f (2)) has slope 4 and therefore equation y 3 = 4( x 2); that is, y = 4 x 5. 2.1.6: Because f ( x ) = 0 · x 2 3 x +7, the slope-predictor for f is m ( a ) = 2 · 0 · a 3 = 3. So the line tangent to the graph of f at (2 , f (2)) has slope 3 and therefore equation y 1 = 3( x 2); that is, y = 3 x + 7. 2.1.7: Because f ( x ) = 2 x 2 3 x + 4, the slope-predictor for f is m ( a ) = 2 · 2 · a 3 = 4 a 3. So the line tangent to the graph of f at (2 , f (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 ) = 2 · ( 1) · a 3 = 2 a 3. So the line tangent to the graph of f at (2 , f (2)) has slope 7 and therefore equation y + 5 = 7( x 2); that is, y = 7 x + 9. 2.1.9: Because f ( x ) = 2 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 , f (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 , f (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 , f (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
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 , f (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 , f (2)) has slope m (2) = 24 and therefore equation y 48 = 24( x 2); that is, y = 24 x .