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- 2.3 - Figure - The parabola y = (x - 2)² (4,029 bytes)
1: ...at is used in [[2.3 Quadratic expressions|section 2.3]] and is written in the programming language [h...
14: t2 := arctime (arclength p -7bp) of p;
15: draw subpath(t1,t2) of p;
17: dashed dashpattern(off 12/9 bp on 9/9 bp);
18: draw subpath(t2,length p) of p - 4.1 - Figure - The equation (x - 1)² + (y - 2)² = 9 (3,564 bytes)
14: t2 := arctime (arclength p -5bp) of p;
15: draw subpath(t1,t2) of p withpen pencircle scaled 0.4pt;
18: draw subpath(t2,length p) of p
25: t2 := arctime (arclength p -7bp) of p;
26: draw subpath(t1,t2) of p; - 4.1 - Figure - The equation x² + (y - 1)² = 1 (3,564 bytes)
14: t2 := arctime (arclength p -5bp) of p;
15: draw subpath(t1,t2) of p withpen pencircle scaled 0.4pt;
18: draw subpath(t2,length p) of p
25: t2 := arctime (arclength p -7bp) of p;
26: draw subpath(t1,t2) of p; - 4.1 - Figure - The equation (x + 1)² + (y - 3)² = 5 (3,572 bytes)
14: t2 := arctime (arclength p -5bp) of p;
15: draw subpath(t1,t2) of p withpen pencircle scaled 0.4pt;
18: draw subpath(t2,length p) of p
25: t2 := arctime (arclength p -7bp) of p;
26: draw subpath(t1,t2) of p; - 4.1 - Figure - The equation (x - 4)² + y² = 13 (3,715 bytes)
14: t2 := arctime (arclength p -5bp) of p;
15: draw subpath(t1,t2) of p withpen pencircle scaled 0.4pt;
18: draw subpath(t2,length p) of p
25: t2 := arctime (arclength p -7bp) of p;
26: draw subpath(t1,t2) of p; - 4.1 - Figure - The equation (x - 3)² + (y - 4)² = 20 (3,857 bytes)
14: t2 := arctime (arclength p -5bp) of p;
15: draw subpath(t1,t2) of p withpen pencircle scaled 0.4pt;
18: draw subpath(t2,length p) of p
25: t2 := arctime (arclength p -7bp) of p;
26: draw subpath(t1,t2) of p; - 2.3 - Figure - The graph of f(x) = (x - 1)² + 2 (3,605 bytes)
1: ...r 2.3:8|the answer]] to [[2.3 Exercises|exercise 2.3:8]] and is written in the programming language ...
14: t2 := arctime (arclength p -7bp) of p;
15: draw subpath(t1,t2) of p;
17: dashed dashpattern(off 12/9 bp on 9/9 bp);
18: draw subpath(t2,length p) of p - 4.1 - Figure - The circle (x - 1)² + (y - 2)² = 3 (3,320 bytes)
14: t2 := arctime (arclength p -7bp) of p;
15: draw subpath(t1,t2) of p;
17: dashed dashpattern(off 12/9 bp on 9/9 bp);
18: draw subpath(t2,length p) of p
19: dashed dashpattern(off 12/9 bp on 9/9 bp); - 4.1 - Figure - The circle (3x - 1)² + (3y + 7)² = 10 (3,328 bytes)
14: t2 := arctime (arclength p -7bp) of p;
15: draw subpath(t1,t2) of p;
17: dashed dashpattern(off 12/9 bp on 9/9 bp);
18: draw subpath(t2,length p) of p
19: dashed dashpattern(off 12/9 bp on 9/9 bp);
Page text matches
- 2.3 Quadratic expressions (12,224 bytes)
1: __NOTOC__
4: {{Selected tab|[[2.3 Quadratic expressions|Theory]]}}
5: {{Not selected tab|[[2.3 Exercises|Exercises]]}}
32: {{Displayed math||<math>x^2+px+q=0</math>}}
38: The equation <math>x^2=a</math> where <math>a</math> is a positive numbe... - 4.1 Angles and circles (13,116 bytes)
1: __NOTOC__
39: ...ns, since the circumference of a circle is <math>2\pi r</math>, where <math>r</math> is the radius o...
44: ...te revolution is <math>360^\circ</math> or <math>2\pi</math> radians which means
46: &1^\circ = \frac{1}{360} \cdot 2\pi\ \mbox{ radians }
48: &1\ \mbox{ radian } = \frac{1}{2\pi} \cdot 360^\circ - Answer 2.3:8 (256 bytes)
3: ...dth="50%" |{{:2.3 - Figure - The graph of f(x) = x² + 1}}
5: ...{{:2.3 - Figure - The graph of f(x) = (x - 1)² + 2}}
8: ...dth="50%" |{{:2.3 - Figure - The graph of f(x) = x² - 6x + 11}} - Solution 2.1:2a (671 bytes)
4: ...x\cdot x-x\cdot 5- 4\cdot x-4\cdot (-5)-(3x \cdot 2x-3x\cdot 3)\\
5: &= x^2-5x-4x+20-(6x^2-9x)\\
6: &=x^2-5x-4x+20-6x^2+9x\,\textrm{.}
9: Then, gather together ''x''²-, ''x''- and the constant terms and simplify
12: ...tom{(x-4)(x-5)-3x(2x-3)}&= (x^2-6x^2)+(-5x-4x+9x)+20 \\ - Solution 2.1:4a (907 bytes)
3: ...box[#FFEEAA;,1.5pt]{\strut 2\cdot 3x^{2}-2\cdot x+2\cdot 5}\,\textrm{.}}</math>}}
5: Then, do the same for 2 from the first bracket
8: ...box[#FFEEAA;,1.5pt]{\strut 2\cdot 3x^{2}-2\cdot x+2\cdot 5}\,\textrm{.}</math>}}
10: Now, collect together ''x''³-, ''x''²-, ''x''- and the constant terms
13: <math>3x^{3}+(-1+6)x^{2}+(5-2)x+10=3x^{3}+5x^{2}+3x+10\,\textrm{.}</math>}} - Solution 2.1:4b (1,488 bytes)
3: {{Displayed math||<math>(1+x+x^{2}+x^{3})(2-x+x^{2}+x^{4})</math>}}
8: &(1+x+x^{2}+x^{3})(2-x+x^{2}+x^{4})\\[3pt]
9: ...ot 2+1\cdot (-x)+1\cdot x^{2}+1\cdot x^{4}+x\cdot 2+x\cdot (-x) \\
10: ...+x^{2}\cdot 2+x^{2}\cdot (-x)+x^{2}\cdot x^{2}+x^{2}\cdot x^{4} \\
11: ...quad{}+x^{3}\cdot 2+x^{3}\cdot (-x)+x^{3}\cdot x^{2}+x^{3}\cdot x^{4}\,\textrm{.} - Solution 2.1:4c (861 bytes)
1: ... brackets together give terms in ''x''¹ and ''x''².
5: ...7x^{2}-x^{4}) = \cdots + \underline{x\cdot 1\cdot 2} + \cdots</math>}}
7: ...coefficient in front of ''x'' is <math>1\cdot 2 = 2\,</math>.
9: As for ''x''², we also have only one possible combination
11: ...x^{2}-x^{4}) = \cdots + \underline{x\cdot 3x\cdot 2} + \cdots</math>}} - Solution 2.2:3c (2,202 bytes)
7: &= \frac{2}{(x-1)(x+1)}\,\textrm{.}
12: ...||<math>\frac{2}{(x-1)(x+1)}\bigl(x^{2}+\tfrac{1}{2}\bigr) = \frac{6x-1}{3(x-1)}\,\textrm{.}</math>}}
17: ...ed math||<math>\frac{2}{x+1}\bigl(x^{2}+\tfrac{1}{2}\bigr) = \frac{6x-1}{3}\,\textrm{.}</math>}}
21: {{Displayed math||<math>6\bigl(x^{2}+\tfrac{1}{2}\bigr) = (6x-1)(x+1)\,\textrm{.}</math>}}
25: {{Displayed math||<math>6x^{2}+3=6x^{2}+5x-1\,\textrm{.}</math>}} - Solution 2.3:2a (1,139 bytes)
1: ...ond order equation by combining together the ''x''²- and ''x''-terms by completing the square to obta...
5: ...m{()}}+3 = \underline{(x-2)^{2}-2^{2}}+3 = (x-2)^{2}-1\,\textrm{,}</math>}}
9: {{Displayed math||<math>(x-2)^{2}-1 = 0</math>}}
13: :*<math>x-2=\sqrt{1}=1\,,\ </math> i.e. <math>x=2+1=3\,,</math>
15: ...*<math>x-2=-\sqrt{1}=-1\,,\ </math> i.e. <math>x=2-1=1\,\textrm{.}</math> - Solution 2.3:2d (1,373 bytes)
1: ...lized form (i.e. the coefficient in front of ''x''² is 1) by dividing both sides by 4,
3: {{Displayed math||<math>x^{2}-7x+\frac{13}{4}=0\,\textrm{.}</math>}}
8: x^{2}-7x+\frac{13}{4}
9: ...-\frac{7}{2}\Bigr)^{2} - \Bigl(\frac{7}{2}\Bigr)^{2} + \frac{13}{4}\\[5pt]
10: &= \Bigl(x-\frac{7}{2}\Bigr)^{2} - \frac{49}{4} + \frac{13}{4}\\[5pt] - Solution 2.3:7a (328 bytes)
1: ...polynomial's largest value is equal to <math>1-0^{2}=1\,</math>. - Solution 2.3:8a (632 bytes)
1: ...he left and, compared with that curve, <math>y=x^{2}+1</math>
6: |align="center"|[[Image:2_3_8_a-1.gif|center]]
8: |align="center"|[[Image:2_3_8_a-2.gif|center]]
10: ...<small>The graph of ''f''(''x'') = ''x''²</small>
12: ...<small>The graph of ''f''(''x'') = ''x''² + 1</small> - Solution 2.3:8b (774 bytes)
1: ...2</math> which is a parabola with a minimum at (0,2) and is sketched further down. Compared with that...
2: ...one unit to the right compared with <math>y=x^{2}+2</math>.
6: |align="center"|[[Image:2_3_8_b-1.gif|center]]
8: |align="center"|[[Image:2_3_8_b-2.gif|center]]
10: ...h of ''f''(''x'') = ''x''² + 2</small> - Solution 2.3:8c (706 bytes)
3: ...^{2}-6x+11 = (x-3)^{2} - 3^{2} + 11 = (x-3)^{2} + 2,</math>}}
5: ...ath> is the same curve as the parabola <math>y=x^{2}</math>, but shifted two units up and three units...
9: |align="center"|[[Image:2_3_8_c-1.gif|center]]
11: |align="center"|[[Image:2_3_8_c-2.gif|center]]
13: ...<small>The graph of ''f''(''x'') = ''x''²</small> - Solution 2.3:10a (623 bytes)
1: ...> define the region above the parabola <math>y=x^{2}</math> and under the line <math>y=1</math>, resp...
4: |align="center"|[[Image:2_3_10_a-1.gif|center]]
6: |align="center"|[[Image:2_3_10_a-2.gif|center]]
8: ...enter"|<small>The region ''y'' ≥ ''x''²</small>
16: [[Image:2_3_10_a2.gif|center]] - Solution 2.3:10b (2,392 bytes)
1: The inequality <math>y\le 1-x^{2}</math> defines the area under and on the curve
2: ...area under and on the straight line <math>y=x/2+3/2</math>.
6: |align="center"|[[Image:2_3_10_b1-1.gif|center]]
8: |align="center"|[[Image:2_3_10_b1-2.gif|center]]
10: ...all>The region y ≤ 1 - ''x''²</small> - Solution 2.3:10c (1,132 bytes)
1: ...e reversed roles, so the inequality <math>x\ge y^{2}</math> defines the same type of parabolic region...
5: |align="center"|[[Image:2_3_10_c1-1.gif|center]]
7: |align="center"|[[Image:2_3_10_c1-2.gif|center]]
11: ...enter"|<small>The region ''x'' ≥ ''y''²</small>
19: |align="center"|[[Image:2_3_10_c2.gif|center]] - Solution 2.3:10d (829 bytes)
1: ...es define the region above the parabola <math>y=x^2</math> and the region below the line <math>y=x</m...
5: |align="center"|[[Image:2_3_10_d1-1.gif|center]]
7: |align="center"|[[Image:2_3_10_d1-2.gif|center]]
9: |align="center"|<small>The region ''x''² ≤ ''y''</small>
19: |align="center"|[[Image:2_3_10_d2.gif|center]] - Solution 3.2:6 (1,830 bytes)
5: ...||<math>\bigl(\sqrt{x+1}+\sqrt{x+5}\,\bigr)^2 = 4^2</math>}}
9: ... + 2\sqrt{x+1}\sqrt{x+5} + \bigl(\sqrt{x+5}\bigr)^2 = 16</math>}}
13: {{Displayed math||<math>x+1+2\sqrt{x+1}\sqrt{x+5}+x+5=16\,\textrm{.}</math>}}
17: {{Displayed math||<math>2\sqrt{x+1}\sqrt{x+5}=-2x+10</math>}}
21: ...th>\bigl(2\sqrt{x+1}\sqrt{x+5}\bigr)^2 = (-2x+10)^2</math>}} - Answer 4.1:6 (529 bytes)
5: ||{{:4.1 - Figure - The circle x² + y² = 9}}
12: ||{{:4.1 - Figure - The circle (x - 1)² + (y - 2)² = 3}}
14: ... with radius √3 and<br> centre at the point (1, 2)</small>
20: ...{{:4.1 - Figure - The circle (3x - 1)² + (3y + 7)² = 10}}
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