Graphing - Part II

Prepared for Gina. L

Questions: Please fully identify and sketch the following functions

Function A : y = 2(x+2)2

Function B : y = 2x3 − 8x2 − 14x + 20

Function C : y = x3

Step 1 - Domain and Range (we had cover this in pervious lecture)

Function A:  y = 2(x+2)2

Domain (∞, ∞)

Range [0, ∞)

Function B:  y = 2x3 − 8x2 − 14x + 20

Domain (∞, ∞)

Range (∞, ∞)

Function C: y = x3

Domain (∞, ∞)

Range (∞, ∞)

Step 2 - Even / Odd function (we had cover this in pervious lecture)

Function A:  y = 2(x+2)2

y(x) = 2(x+2)2

= 2(x2 + 2x + 4)

y(x) = 2((x)+2)2

= 2(−x+2)2

= 2(x2 2x + 4)

y(x) = 2(x2 + 2x + 4) 2(x2 2x + 4) = y(x)

y(x) = 2(x2 + 2x + 4)    2(x2 2x + 4) = y(x)

Therefore, function A, y = 2(x+2) is not even/odd function, its neither.

Function B:  y = 2x3 − 8x2 − 14x + 20

y(x) = 2x3 − 8x2 − 14x + 20

y(x) =2(−x)3 − 8(−x)2 − 14(−x) + 20

= −2x3 − 8x2 + 14x + 20

y(x) = 2x3 − 8x2 − 14x + 20 −2x3 − 8x2 + 14x + 20 = y(x)

y(x) = 2x3 − 8x2 − 14x + 20    2x3 + 8x2 − 14x − 20 = y(x)

Therefore, function B, y = 2x3 − 8x2 − 14x + 20  is not even/odd function, its neither.

Function C: y = x3

y(x) = x3

y(−x) = (−x)3

= −x3

y(x) = x3 = −(−x3) = −y(−x)

Therefore, function C, y = x is an odd function.

Step 3. Y-Intercept Point

To find y - intercept point of any function, we have to set x = 0, and get the y value.

Remember, in the case of function, we can only have ONE y-intercept point (one input, one output result)

Function A:  y = 2(x+2)2

set x = 0

we get y = 2(0+2)2

y = 2(2)

y = 8

Therefore, the y-int point for function A is (0, 8)

Function B:  y = 2x3 − 8x2 − 14 x + 20

set x = 0,

we get y = 2(0)3 − 8(0)2 − 14(0) + 20

y = 2(0) − 8(0) − 14(0) + 20

y  = 20

Therefore, the y-int point for function B is (0, 20)

Function C: y = x3

set x = 0

we get y = (0)3

y = 0

Therefore, the y-int point for function C is (0, 0)

Step 4 . Zeros (X - Intercept Points)

To find x - intercept points, we set y = 0,

notice, we can have zero, one or more than one zeros in a function (same output, could come from many different inputs)

Function A:  y = 2(x+2)2

set y = 0

we get 0 = 2(x+2)2

0 = (x+2)2

0 = (x+2)

x = −2

Therefore, the zeros (x-int point) for function A is (−2, 0)

Function B:  y = 2 x3 − 8x2 − 14 x + 20

set y = 0,

we get 0 = 2x3 − 8x2 − 14 x + 20

since we cannot go further on factor, we can try long division to factor it,

plug into x = 1, 2, 3,.. so on to test if y = 0 (find at least one zero on the function).

start from x = 1

y = 2(1)3 − 8(1)2 − 14(1) + 20

= 2 − 8 − 14 + 20

= 2 − 22 + 20

= 0   (Yes, we find the zero!!)

so at x = 1, there exist a zero, so we can said that (x − 1) is one of the factor of y  = 2x3 − 8x2 − 14 x + 20.

by long division method, we write the following... we get 2x2 − 6x − 20, now go back to our factor part left above,

y = 2 x3 − 8x2 − 14 x + 20

= (x − 1)(2x2 − 6 x − 20)

= 2(x − 1)(x2 − 3x − 10)

= 2(x − 1)(x − 5)(x + 2)

set y = 0

0 = 2(x − 1)( x − 5)(x + 2)

x = 1 or x = 5 or x = −2

Therefore, the zeros (x-int points) for function B are (−2, 0), (1, 0) and (5, 0).

Function C: y = x3

set y = 0

we get 0 = x3

x = 0

Therefore, the zeros (x-int points) for function C is (0, 0)

Step 5 . Positive or Negative

Function A:  y = 2(x+2)2

draw a table to help us find out where is the output (y) of function become positive or negative.

the key point is used the zeros (x-int) as a border, from step 4, we get x-int point as (−2, 0)

 Input (x) x < −2 (try x = −3) x > −2 (try x = 0) y = 2(x+2)2 y = 2(−3+2)2    = 2(1) y = 2(0+2)2    = 2(4) Output (y) + Positive + Positive

Function B:  y = 2 x3 − 8x2 − 14 x + 20

from step 4, we get x-int points as (−2, 0), (1, 0) and (5, 0)

 Input (x) x < −2 (try x = −3) −2 < x < 1 (try x = 0) 1 < x < 5 (try x = 2) x > 5 (try x =6) y = 2 x3 − 8x2 − 14 x + 20 y = 2(−3)3 − 8(−3)2 − 14(−3) + 20    = −54 − 32 + 42 + 20    = − 24 y = 2(0)3 − 8(0)2 − 14(0) + 20     = 0 − 0 − 0 + 20    = 20 y = 2(2)3 − 8(2)2 − 14(2) + 20    = 16 − 32 − 28 + 20    = − 24 y = 2(6)3 − 8(6)2 − 14(6) + 20    = 432 − 288 − 84 + 20    = 80 Output (y) − Negative + Positive − Negative + Positive

Function C: y = x3

from step 4, we get x-int point as (0, 0)

 Input (x) x < 0 (try x = −1) x > 0 (try x = 1) y = x3 y = (−1)3   = −1 y = (1)3   = 1 Output (y) − Negative + Positive

Step 6. Asymptotes

Vertical Asymptotes, Horizontal Asymptotes, Oblique Asymptotes

We will cover it later.

Step 7. 1st Order Derivative (y')

Critical Points and Increasing or Decreasing

Function A:  y = 2(x+2)2

y' = dy/dx = d/dx 2(x+2)2

= 4(x + 2)

set dy/dx = 0

0 = 4(x + 2)

x = − 2

when x = − 2, y = 0

the critical point is (− 2, 0)

 Input (x) x < −2 (try x = −3) x > −2 (try x = 0) y' = 4(x + 2) y' = 4(−3 + 2)    = −4 y' = 4(0 + 2)    = 4 Slope (y') − Decreasing + Increasing

Function B:  y = 2 x3 − 8x2 − 14 x + 20

y' = dy/dx = d/dx  2x3 − 8x2 − 14 x + 20

= 6x2 − 16x − 14

set dy/dx = 0

0 = 6x2 − 16x − 14

0 = 2(3x2 − 8x − 7)

0 = 3 x2 − 8x − 7

again, we cannot factor it, however, this time, we don't need to go to long division, we can use quadratic equation to solve it.

a = 3, b = − 8, c = − 7 when x = 3.36, y = 2(3.36)3 − 8(3.36)2 −14(3.36) + 20 = −94.77

and when x = −0.69, y = 2(−0.69)3 − 8(−0.69)2 − 14(−0.69) + 20 = 25.19

the critical point are (3.36, −94.77) and (−0.69, 25.19).

 Input (x) x < −0.69 (try x = −1) −0.69< x < 3.36 (try x = 0) x > 3.36 (try x = 4) y' = 6x2 − 16x − 14 y' = 6(−1)2 − 16(−1) − 14    = 8 y' = 6(0)2 − 16(0) − 14    = − 14 y' = 6(4)2 − 16(4) − 14    = 18 Slope (y') + Increasing − Decreasing + Increasing

Function C: y = x3

y' = dy/dx = d/dx x3

= 3x2

set dy/dx = 0

0 = 3x2

x = 0

when x = 0, y = 0

the critical point is (0, 0)

 Input (x) x < 0 (try x = −1) x > 0 (try x = 1) y' = 3x2 y' = 3(−1)2    = 3 y' = 3(1)2    = 3 Slope (y') + Increasing + Increasing

Step 8. 2nd Order Derivative (y")

Min and Max points, Curve Up or Curve Down, Point of Inflection and Cusp

Function A:  y = 2(x+2)2

Step 8.1 - find 2nd order derivative

y" = d2y/dx2 = d/dx y'

= d/dx 4(x + 2)

= 4

Step 8.2 find min/max points

Notice the critical points (step-7) we get, they may be Min / Max point,

in order to identify them, we MUST test them by 2nd order derivative, from step-7 we get the critical point as (− 2, 0),

sub x = − 2 into y", we get

y" = 4

Since y" = 4 > 0  (positive), it is a MIN point,

therefore we get one MIN point at (− 2, 0)

Step 8.3 Curve Up or Curve Down

y"=4 > 0 for all along the function.

 Input (x) x y" = 4 y" = 4 Change of Slope (y") + Curve Up

P.S. For easier way to remember curve up or down and Min or MAX, you may think like this,

if y" >0 = positive number, think it as a HAPPY face (curve UP), and there is a MIN point (minimum of anger).  if y" < 0 = negative number, think it as a SAD face (curve DOWN), and there is a MAX point (maximum of anger).  Step 8.4 Point of Inflection and Cusp

Since y"≠ 0, there are NO point of inflection or cusp.

Function B:  y = 2x3 − 8x2 − 14x + 20

y" = d2y/dx2 = d/dx y'

= d/dx 3x2 − 8x − 7

= 6x − 8

the critical points from step 7 are (3.36, −94.77) and (−0.69, 25.19), sub them into y".

when x = 3.36

y" = 6 (3.36) − 8 = 12.16 > 0 (positive), so (3.36, −94.77) is a MIN point.

when x = −0.69

y" = 6( −0.69) − 8 = −12.14 < 0 (negative), so (−0.69, 25.19) is a MAX point.

then, we set y" = 0

0 = 6x − 8

x = 8/6

= 1.333...

y = 2(1.33)3 − 8(1.33)2 − 14(1.33) + 20

= − 8.06

 Input (x) x < 1.333... (try x = 0) x > 1.333... (try x = 2) y" = 6x − 8 y" = 6(0) − 8    = − 8 y" = 6(2) − 8    = 4 Change of Slope (y") − Curve Down + Curve Up

At x = 8/6 = 1.333... , it might be a point of inflection or cusp, in order to identify it,

we have to check if it's increasing or decreasing, now by combine the two table, we can easier to see the relationship between curve up/down and increasing/ decreasing.

 Input (x) x < −0.69 (try x = −1) −0.69< x < 1.33.. (try x = 0) 1.33..< x < 3.36 (try x = 2) x > 3.36 (try x = 4) y' = 6x2 − 16x − 14 y' = 6(−1)2 − 16(−1) − 14    = 8 y' = 6(0)2 − 16(0) − 14    = − 14 y' = 6(2)2 − 16(2) − 14    = −22 y' = 6(4)2 − 16(4) − 14    = 18 Slope (y') + Increasing − Decreasing − Decreasing + Increasing y" = 6x − 8 y" = 6( −1) − 8    = − 14 y" = 6(0) − 8    = − 8 y" = 6(2) − 8    = 4 y" = 6(4) − 8    = 16 Change of Slope (y") − Curve Down − Curve Down + Curve Up + Curve Up

curve up/down and increasing/ decreasing.

Take a look at the column are shaded, before x = 1.33 it decreasing, after x = 1.33 it still decreasing, however,

it curve down before x=1.33 but curve up after x=1.33, since the sing of y" change, point (1.33, − 8.06) is a point of inflection.

P.S. Notice if the sign of y" does not change, it is cusp point

Point of Inflection

Decreasing -> Decreasing

Curve Down -> Curve Up

Cusp

Decreasing -> Increasing

Curve Down -> Curve Down

Function C: y = x3

y" =d2y/dx2 = d/dx y'

= d/dx 3x2

= 6x

the critical points from step 7 is (0, 0), sub x = 0 into y".

y" = 6 (0) = 0, we cannot tell it is MIN or MAX (we have four possibility, it could be MIN, MAX, Point of Inflection or Cusp)

To identify which one is the correct one, we have to check is the function curve up or curve down.

set y" = 0

0 = 6x

x = 0

 Input (x) x < 0... (try x = −1) x > 0... (try x = 1) y" = 6x y" = 6(−1)    = − 6 y" = 6(1)    = 6 Change of Slope (y") − Curve Down + Curve Up

combine the two table, we can easier to see the relationship between curve up/down and increasing/ decreasing.

 Input (x) x < 0 (try x = −1) x > 0 (try x = 1) y' = 6x2 y' = 6(−1)2     = 6 y' = 6(1)2     = 6 Slope (y') + Increasing + Increasing y" = 6x y" = 6(−1)    = − 6 y" = 6(1)    = 6 Change of Slope (y") − Curve Down + Curve Up

Take a look before x = 0 it is increasing, after x = 0 it still increasing, however,

it curve down before x=0 but curve up after x=0, since the sing of y" change, point (0,0) is a point of inflection.

Step 9. Sketching

Function A : y = 2(x+2)2

 Domain (−∞, ∞) Range [0, ∞) Even / Odd neither y-int point (0, 8) zeros (x-int point) (−2, 0) critical point is (− 2, 0) local min/max local min at (− 2, 0) curve curve up (−∞, ∞)

As well as the positive/negative region, increasing/decreasing region's information, you shall able to sketch Function.A easily. Additionally, you may wish to check the result by using a powerful math tool - MATLAB (http://www.mathworks.com)

 % Function A y = 2(x+2)^2 plots, simply MATLAB script written by KK. 　 ezplot ('2*(x+2)^2',[-10,10]);   % The domain for x set between -10 to 10 grid;                                          % Add grid to the function title ('Function A: y = 2(x+2)^2');  %Function title ylabel ('Y-axis');                            %Y-axis name xlabel ('X-axis');                            %X-axis name

it generates the following output. Repeat the same procedures as we had done above, now, try to sketch function B and C.

Function B : y = 2x3 − 8x2 − 14x + 20 % Function B 2(x^3) − 8(x^2) − 14x + 20 plots, simply MATLAB script written by KK. 　 ezplot ('2*(x^3) - 8*(x^2) - 14*x + 20',[-4,6]);   % The domain for x set between -4 to 6 grid;                                                                         % Add grid to the function title ('Function B: y = 2(x^3) − 8(x^2) − 14x + 20');  %Function title ylabel ('Y-axis');                                                       %Y-axis name xlabel ('X-axis');                                                       %X-axis name

and with image tool "Data Cursor" in MATLAB, we can trace the function's local max, min on the plots (its just for checking purpose). Function C: y = x3

By online graphing tool. By MATLAB All the functions are now fully sketched.

Good luck and see you later ~ lol

Written by Kevin Tang (Mar 28, 2010)

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