Understanding Amplitude, Midline, and Period from Graph Points
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Amplitude is the vertical distance from the midline to a maximum or minimum point. It can be calculated as half the difference between the maximum and minimum values: [ \text{Amplitude} = \frac{\text{Maximum} - \text{Minimum}}{2} ]
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Midline (D) is the vertical midpoint between the maximum and minimum points: [ D = \frac{\text{Maximum} + \text{Minimum}}{2} ]
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Period (P) is the horizontal length of one full cycle of the wave. It equals twice the horizontal distance between a maximum and the next minimum point: [ P = 2 \times (\text{Horizontal distance between max and min}) ]
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The parameter ( B ) in the sinusoidal function ( f(x) = A \sin(Bx) + D ) or ( f(x) = A \cos(Bx) + D ) relates to the period by: [ B = \frac{360^\circ}{P} ]
Example: Modeling a Cosine Function
Given a curve with:
- Maximum point at ( (1, 3.5) )
- Minimum point at ( (2, 0.5) )
- ( a < 0 ) (amplitude negative)
Calculate:
- Amplitude ( |a| = \frac{3.5 - 0.5}{2} = 1.5 )
- Midline ( c = \frac{3.5 + 0.5}{2} = 2 )
- Period ( P = 2 \times (2 - 1) = 2 )
- ( B = \frac{360}{2} = 180 )
Function form: [ f(t) = -1.5 \cos(180t) + 2 ]
Solving for Specific Values
Number of Solutions for ( f(t) = 2 )
- The horizontal line ( y = 2 ) intersects the graph 6 times within the domain ( t \in [1, 5] ).
Number of Solutions for ( f(t) = 0.5 )
- The line ( y = 0.5 ) intersects the graph 3 times.
Solving ( f(t) = 3 ) for ( t \in [0, 2] )
- Set up the equation: [ 1.5 \cos(180t) + 2 = 3 ]
- Using a graphing calculator, find intersections at:
- ( t_1 = 0.732 )
- ( t_2 = 1.27 )
Solving Inequality ( f(t) \leq 1.5 ) for ( t \in [0, 2] )
- Inequality: [ 1.5 \cos(180t) + 2 \leq 1.5 ]
- Graphing reveals solutions in intervals:
- ( t \in [0, 0.392] )
- ( t \in [1.61, 2] )
Using a Graphing Calculator
- Enter the function ( f(t) = 1.5 \cos(180t) + 2 ) in the graphing tool.
- Adjust window settings:
- ( x_{min} = 0 ), ( x_{max} = 2 )
- ( y_{min} = -1 ), ( y_{max} = 5 )
- Scale appropriately for clear viewing.
- Use intersection features to find exact solution points.
- Change display settings to auto for precise decimal values.
Summary
This tutorial covers how to extract key parameters from sinusoidal graphs and solve related equations and inequalities. It emphasizes the relationship between maximum/minimum points and amplitude, midline, and period, and demonstrates practical graphing techniques to find solutions within specified domains.
For further understanding of trigonometric functions, you may find these resources helpful:
- Mastering Trigonometric Identities, Equations, and the CAST Diagram
- Solving Square Area, Function Inverses, and Rational Functions Explained
- Le Proprietà Fondamentali delle Funzioni Trigonometriche
- Solving Varying Angle Problems Using Sine and Cosine Laws
- Calculating Arc Length, Triangle, and Sector Areas with Theta
normally on the graph you are given maximum and minimum points use the coordinates of Maximum
and a minimum of points we can figure out amplitude midline also the period we
know amplitude is the distance between midline and maximum
point in other words it's half the distance between maximum and minimum
points amplitude equals maximum minus minimum over two mid line is the the vertical middle
point between maximum and minimum this D equals maximum plus minimum over 2 we
know the period equals the distance between two consecutive maximum points or two
consecutive minimum points therefore the horizontal distance between maximum and minimum
points will be half of the period so we get period equals double the horizontal distance between maximum and minimum
points using this formula P equal 360° over B we can figure out to this b equals
360° over period in IB AO level no horizontal translation so in this video we only
discuss f ofx = a sin BX + D or F ofx = A cosine BX + D we know this D is a vertical trans translation yal D it's
the equalent of principal axis absolute value of a is the
amplitude this B we can use 360° over period to figure out go to question three the curve shown can be
modeled by the function f of t equal a cosine bt+ c t greater than or equal
to1 less than or equal to 5 a is a less than zero the curve has a maximum point at 1
3.5 and a minimum point at 20.5 amplitude will be absolute value of a is half the distance between maximum
and minimum points so amplitude equals
3.05 minus 0.5 over 2 equals 1.5 since a is negative a a =
1.5 find the value of c c is the midline it's the vertical midpoint between maximum and
minimum this is C equals 3.5 + 0.5 over 2 equal 2 find the value of B this b equals 360 over
period let's check the period period equals double the horizontal distance between these two consecutive maximum
and minimum points 2 - 1 so Period = 2 B = 360 / 2 =
180 question four when F of T = 2 find the number of four Solutions F of tal 2 let's draw this
line basically it's mid line since the t is a greater than or equal to -1 less than or equal to
5 how many intersections we have on the graph will be the number of solutions one 2 3 4
five 6 answer is six question five when F of T equals
0.05 five find the number of solutions 0.5 so here we have a one two three
intersections so answer is a three question six solve F of T = 3 for T greater than or equal to Z
less than or equal to two we are given this domain so when you use graphing calculator to solve you have to set up
X minimum equals zero X maximum equals 2 x scale will be period over four since we already figure out since we already
figure out a c and a b f of T equals
1.5 coine 180 t + 2 we will set up equation like this 1.5 coine
180t + 2 equals 3 we will solve this equation by graphing this still use the graph on scratch Pad clear the
history menu action delete
all press tab F1 of X = 1.5 H
sign 180x plus two enter Press menu to change the window
setting four enter Max minimum is zero maximum
two scale 2 / 4 y minimum let's check this uh graph where
minimum it's a netive 1 y maximum we can put five scale let's put one then
enter press tab again FS2 of x = 3
enter we need to figure out these two intersections Manu analyze the graph intersection four
control menu recent intersection so we get T1 equal
0.732 T2 equals 1.27 make sure you change the setting for the graph by doing
menu setting display digits into Auto
and automatically label points by graphing T1 equals
0.732 T2 equals 1.72 question 7 solve F of a t less than
or equal to 1.5 for T greater than or equal to Z less than or equal to 2 we will set up
inquality 1.5 coine
180t + 2 less than or equal to 1.5 we will solve this inequality by graphing go back to graph for on scratch
pack click on Tab upper Arrow F2 of x
equals let's put 1.5 enter we need to figure out these two
intersections menu six four then control menu
recent intersection we got two Roots
T1 equal 0.392 as a 36 F T2 = 1.61 from this graph we could
see this part of our graph and uh this part of graph the value of f of a t less than
1.5 so the solution to this inequality will be T greater than or equal to Z less than or equal to
0.392 or t greater than or equal to 1.61 less than or equal to
2
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