logging in or signing up Graphs & Graphical Transformations Zarrata Download Post to : URL : Related Presentations : Share Add to Flag Embed Email Send to Blogs and Networks Add to Channel Uploaded from authorPOINT lite Insert YouTube videos in PowerPont slides with aS Desktop Copy embed code: (To copy code, click on the text box) Embed: URL: Thumbnail: WordPress Embed Customize Embed The presentation is successfully added In Your Favorites. Views: 1203 Category: Education License: Some Rights Reserved Like it (1) Dislike it (0) Added: September 26, 2010 This Presentation is Public Favorites: 0 Presentation Description No description available. Comments Posting comment... Premium member Presentation Transcript Pre-Calculus : Pre-Calculus An Introduction To Graphs -Rishabh Dev Zarrata.com Slide 2: Zarrata.com Slide 3: Algebraic Functions π /4 (odd : symmetric about origin) y=x y=x 2 (even : symmetric about the y-axis) y=x 3 y=1/x (Rectangular Hyperbola : a=b, ε = √ 2) Zarrata.com The Modulus Function : The Modulus Function x for x≥0 -x for x≤0 √ x² |x| (Even : Symmetric about y-axis) Zarrata.com Slide 5: x = [x] + {x} (Integral part of x) (fractional part of x) The floor function (A ‘sawtooth’ function) 1 2 3 4 1 1 2 3 1 2 3 Zarrata.com Slide 6: A very important function in computer science, It gives the sign of its argument The Signum Function 1 -1 y=sgn(x) Zarrata.com Geometric Curves : The Conic Sections : Circle: x2 + y2 = r2 Ellipse: , Parabola: y2 = 4ax, x2 = 4ay Hyperbola: , Rectangular Hyperbola: xy = c2 General Equations: Geometric Curves : The Conic Sections Ax + Bxy + Cy + Dx + Ey + F = 0 with A, B, C not all zero,then: if B − 4AC < 0, the equation represents an ellipse (or no curve); if A = C and B = 0, the equation represents a circle; if B − 4AC = 0, the equation represents a parabola; if B − 4AC > 0, the equation represents a hyperbola; if we also have A + C = 0, the equation represents a rectangular hyperbola. 2 2 2 2 2 Zarrata.com Slide 8: The Trigonometric Curves y=sinx y=cosx π/2 π -π/2 π 3π/2 -3π/2 Zarrata.com Slide 9: π/2 π -π/2 π 3π/2 -3π/2 π/4 1 y=tanx y=cotx Zarrata.com Pre-Calculus : Pre-Calculus Graphical Transformations -By Rishabh Dev Zarrata.com Slide 11: f(x) f(x)-a Shift the graph downward through ‘a’ units f(x) f(x)+a Shift the graph upward through ‘a’ units y=|x| y=|x|+a y=|x|-a a a Zarrata.com Slide 12: f(x) f(x+a) Shift the graph ‘a’ units left f(x) f(x-a) Shift the graph ‘a’ units right a a y=|x| y=|x-a| y=|x+a| Zarrata.com Slide 13: Q1)Plot y=x -4 & check weather it is even or odd? Q2)Show the region enclosed by y=x , y=(x+1) , y=(x-1) ? 2 2 2 2 Zarrata.com Slide 14: (x-1) (x+1) 2 2 y= -4 y=x 2 -4 Solutions: Zarrata.com Slide 15: f(x) f(ax) Shrink the graph ‘a’ times along the x-axis f(x) f(x/a) Stretch the graph ‘a’ times along the x-axis pi/2 -pi/2 1 -1 y=sin (2x) -1 y=sin x -1 y=sin (x/2) -1 Zarrata.com Slide 16: f(x) f(-x) Mirror image about the y-axis y=e y=e x -x Zarrata.com Slide 17: f(x) -f(x) Mirror image about x-axis y=lnx y=-lnx Zarrata.com Slide 18: f(x) -f(-x) (1)Image about y (2)Image about x y=e y=e x -x (2) (1) y= -e -x (OR THE OTHER WAY ROUND) Zarrata.com Slide 19: f(x) |f(x)| Mirror image about x-axis of portion lying below the x-axis (Remove that portion after taking mirror image) 1 0 y=lnx 1 Zarrata.com Slide 20: f(x) f|x| Mirror image about y-axis for portion x≥0; neglect portion with x<0 (neglected) y=e |x| 1 y=e x Zarrata.com Slide 21: f(x) |f|x|| (1) Mirror image about x-axis of portion lying below the x-axis (2)Mirror image about y-axis for portion x≥0; neglect portion with x<0 (OR THE OTHER WAY ROUND) (neglected) y=f(x) y=|f(x)| y=| f|x| | Zarrata.com Slide 22: y=f(x) |y|=f(x) (1)Neglect portion below x-axis (2)Plot remaining part and its mirror image in the x-axis (neglected) y=sinx |y|=sinx Zarrata.com Slide 23: Q1)Plot |y|=|log|x|| & find its domain & range? Q2)Plot the curve |x|+|y|=1 and find the area enclosed by it? Zarrata.com Slide 24: Solution (1): y=lnx y=ln|x| y=|ln|x|| |y|=|ln|x|| [Domain : R-{0}, Range : (-∞ , +∞ )] Zarrata.com Slide 25: Solution (2): 1 1 y=1-x y=1-|x| 1 1 1 1 -1 -1 -1 Hence, Required Area = 2 sq units; Since, side length=√ 2 Zarrata.com Slide 26: f(x) [f(x)] 1 -1 0 π/2 π 3π/2 -π/2 π (1)Mark the intervals of unit length with integers as end points on y-axis & mark the corresponding intervals on the x-axis (2)Plot [f(x)] for these marked intervals) y=[sinx] Zarrata.com Slide 27: f(x) f[x] y=e x 1 2 3 -1 -2 Plot lines for all integral x; f[x] includes values n≤ x<n+1 where n is an integer y=e [x] Zarrata.com Slide 28: f(x) f{x} 1 O a a 1 2 3 4 y=f{x} Consider only the portion of graph b/w (0,1]; repeat it for other integers Zarrata.com Slide 29: f(x) {f(x)} 0 1 2 3 Plot integral values of y; draw vertical lines for points of intersection with f(x) & translate the graph b/w 0≤y<1 a b c -a -b -c a b c -a -b -c y={f(x)} Zarrata.com Slide 30: Relation b/w a Function & its Inverse y=x 3 y=x 1/3 y=e x y=lnx 1 1 (mirror image about y=x) Zarrata.com You do not have the permission to view this presentation. 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Graphs & Graphical Transformations Zarrata Download Post to : URL : Related Presentations : Share Add to Flag Embed Email Send to Blogs and Networks Add to Channel Uploaded from authorPOINT lite Insert YouTube videos in PowerPont slides with aS Desktop Copy embed code: (To copy code, click on the text box) Embed: URL: Thumbnail: WordPress Embed Customize Embed The presentation is successfully added In Your Favorites. Views: 1203 Category: Education License: Some Rights Reserved Like it (1) Dislike it (0) Added: September 26, 2010 This Presentation is Public Favorites: 0 Presentation Description No description available. Comments Posting comment... Premium member Presentation Transcript Pre-Calculus : Pre-Calculus An Introduction To Graphs -Rishabh Dev Zarrata.com Slide 2: Zarrata.com Slide 3: Algebraic Functions π /4 (odd : symmetric about origin) y=x y=x 2 (even : symmetric about the y-axis) y=x 3 y=1/x (Rectangular Hyperbola : a=b, ε = √ 2) Zarrata.com The Modulus Function : The Modulus Function x for x≥0 -x for x≤0 √ x² |x| (Even : Symmetric about y-axis) Zarrata.com Slide 5: x = [x] + {x} (Integral part of x) (fractional part of x) The floor function (A ‘sawtooth’ function) 1 2 3 4 1 1 2 3 1 2 3 Zarrata.com Slide 6: A very important function in computer science, It gives the sign of its argument The Signum Function 1 -1 y=sgn(x) Zarrata.com Geometric Curves : The Conic Sections : Circle: x2 + y2 = r2 Ellipse: , Parabola: y2 = 4ax, x2 = 4ay Hyperbola: , Rectangular Hyperbola: xy = c2 General Equations: Geometric Curves : The Conic Sections Ax + Bxy + Cy + Dx + Ey + F = 0 with A, B, C not all zero,then: if B − 4AC < 0, the equation represents an ellipse (or no curve); if A = C and B = 0, the equation represents a circle; if B − 4AC = 0, the equation represents a parabola; if B − 4AC > 0, the equation represents a hyperbola; if we also have A + C = 0, the equation represents a rectangular hyperbola. 2 2 2 2 2 Zarrata.com Slide 8: The Trigonometric Curves y=sinx y=cosx π/2 π -π/2 π 3π/2 -3π/2 Zarrata.com Slide 9: π/2 π -π/2 π 3π/2 -3π/2 π/4 1 y=tanx y=cotx Zarrata.com Pre-Calculus : Pre-Calculus Graphical Transformations -By Rishabh Dev Zarrata.com Slide 11: f(x) f(x)-a Shift the graph downward through ‘a’ units f(x) f(x)+a Shift the graph upward through ‘a’ units y=|x| y=|x|+a y=|x|-a a a Zarrata.com Slide 12: f(x) f(x+a) Shift the graph ‘a’ units left f(x) f(x-a) Shift the graph ‘a’ units right a a y=|x| y=|x-a| y=|x+a| Zarrata.com Slide 13: Q1)Plot y=x -4 & check weather it is even or odd? Q2)Show the region enclosed by y=x , y=(x+1) , y=(x-1) ? 2 2 2 2 Zarrata.com Slide 14: (x-1) (x+1) 2 2 y= -4 y=x 2 -4 Solutions: Zarrata.com Slide 15: f(x) f(ax) Shrink the graph ‘a’ times along the x-axis f(x) f(x/a) Stretch the graph ‘a’ times along the x-axis pi/2 -pi/2 1 -1 y=sin (2x) -1 y=sin x -1 y=sin (x/2) -1 Zarrata.com Slide 16: f(x) f(-x) Mirror image about the y-axis y=e y=e x -x Zarrata.com Slide 17: f(x) -f(x) Mirror image about x-axis y=lnx y=-lnx Zarrata.com Slide 18: f(x) -f(-x) (1)Image about y (2)Image about x y=e y=e x -x (2) (1) y= -e -x (OR THE OTHER WAY ROUND) Zarrata.com Slide 19: f(x) |f(x)| Mirror image about x-axis of portion lying below the x-axis (Remove that portion after taking mirror image) 1 0 y=lnx 1 Zarrata.com Slide 20: f(x) f|x| Mirror image about y-axis for portion x≥0; neglect portion with x<0 (neglected) y=e |x| 1 y=e x Zarrata.com Slide 21: f(x) |f|x|| (1) Mirror image about x-axis of portion lying below the x-axis (2)Mirror image about y-axis for portion x≥0; neglect portion with x<0 (OR THE OTHER WAY ROUND) (neglected) y=f(x) y=|f(x)| y=| f|x| | Zarrata.com Slide 22: y=f(x) |y|=f(x) (1)Neglect portion below x-axis (2)Plot remaining part and its mirror image in the x-axis (neglected) y=sinx |y|=sinx Zarrata.com Slide 23: Q1)Plot |y|=|log|x|| & find its domain & range? Q2)Plot the curve |x|+|y|=1 and find the area enclosed by it? Zarrata.com Slide 24: Solution (1): y=lnx y=ln|x| y=|ln|x|| |y|=|ln|x|| [Domain : R-{0}, Range : (-∞ , +∞ )] Zarrata.com Slide 25: Solution (2): 1 1 y=1-x y=1-|x| 1 1 1 1 -1 -1 -1 Hence, Required Area = 2 sq units; Since, side length=√ 2 Zarrata.com Slide 26: f(x) [f(x)] 1 -1 0 π/2 π 3π/2 -π/2 π (1)Mark the intervals of unit length with integers as end points on y-axis & mark the corresponding intervals on the x-axis (2)Plot [f(x)] for these marked intervals) y=[sinx] Zarrata.com Slide 27: f(x) f[x] y=e x 1 2 3 -1 -2 Plot lines for all integral x; f[x] includes values n≤ x<n+1 where n is an integer y=e [x] Zarrata.com Slide 28: f(x) f{x} 1 O a a 1 2 3 4 y=f{x} Consider only the portion of graph b/w (0,1]; repeat it for other integers Zarrata.com Slide 29: f(x) {f(x)} 0 1 2 3 Plot integral values of y; draw vertical lines for points of intersection with f(x) & translate the graph b/w 0≤y<1 a b c -a -b -c a b c -a -b -c y={f(x)} Zarrata.com Slide 30: Relation b/w a Function & its Inverse y=x 3 y=x 1/3 y=e x y=lnx 1 1 (mirror image about y=x) Zarrata.com