How do you graph y=4cos(2x-1)+3y=4cos(2x1)+3?

1 Answer
Apr 1, 2018

So features of graph

Amp:4
Period: piπ
Max:7
Min:-1

graph{4cos(2x-1)+3 [-10, 10, -5, 5]}

Explanation:

Translation form of sine is

Asin[B(x-C)]+DAsin[B(xC)]+D

A~ Vertical stretch/Amp, y values get stretched by A
B~ Horizontal strech/Period, x values get stretched by 1/B1B
C~ Horizontal translation/Phase shift, x values move over by C
D~ Vertical translation, y values up by D

So we know that if we put it in translation form it looks like

4cos[2(x-1/2)]+34cos[2(x12)]+3

So we know that the original sin(x)sin(x) has these features

Amp:1
Period:2pi2π
Max: 1
Min: -1

graph{sin(x) [-8.89, 8.886, -4.446, 4.446]}

So 4cos(x)4cos(x) means the Amplitude becomes 4 meaning the max is 4 and min -4

So 4cos(2x)4cos(2x) means the period halves becoming piπ

So 4cos[2(x-1/2)]4cos[2(x12)] means the origin moves over by 1/212

4cos[2(x-1/2)]+34cos[2(x12)]+3 means all the y values move up by 3 meaning the max is 7 and the min -1

So features of graph

Amp:4
Period: piπ
Max:7
Min:-1

graph{4cos(2x-1)+3 [-10, 10, -5, 5]}