Question

pre-calculus pre-assessment snap color theme give the range of the function: ( y = 5sin(4x)+1 ) 1.

Explanation:

Step1: Determine the range of the sine function

The range of the basic sine function \(y = \sin(u)\) is \([- 1,1]\). Here \(u = 4x\), and the function \(y=\sin(4x)\) still has a range of \([-1,1]\).

Step2: Consider the vertical stretch

For the function \(y = 5\sin(4x)\), when we multiply \(\sin(4x)\) by \(5\), the range of \(y = 5\sin(4x)\) is obtained by multiplying each value in the range of \(\sin(4x)\) by \(5\). Using the property \(a\cdot f(x)\) where \(a = 5\) and \(f(x)=\sin(4x)\), if \(y_1=\sin(4x)\) and \(y_2 = 5y_1\), then when \(y_1=-1\), \(y_2=-5\) and when \(y_1 = 1\), \(y_2 = 5\). So the range of \(y = 5\sin(4x)\) is \([-5,5]\).

Step3: Consider the vertical shift

For the function \(y=5\sin(4x)+1\), we use the property \(y = f(x)+k\) where \(f(x)=5\sin(4x)\) and \(k = 1\). If \(y_3=f(x)\) and \(y_4=y_3 + 1\), then when \(y_3=-5\), \(y_4=-5 + 1=-4\) and when \(y_3=5\), \(y_4=5 + 1=6\).

Answer:

The range of the function \(y = 5\sin(4x)+1\) is \([-4,6]\)