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example 1 graph the function $y = -\\frac{1}{2}\\cos x$.

Question

example 1
graph the function $y = -\frac{1}{2}\cos x$.

Explanation:

Step1: Determine the amplitude

The amplitude of \(y = A\cos x\) is \(|A|\). For \(y =-\frac{1}{2}\cos x\), \(A =-\frac{1}{2}\), so the amplitude \(|A|=\frac{1}{2}\).

Step2: Analyze the reflection

The negative sign in \(y =-\frac{1}{2}\cos x\) reflects the graph of \(y=\cos x\) about the \(x -\)axis.

Step3: Plot key points

For \(y = \cos x\), key points are \((0,1)\), \((\frac{\pi}{2},0)\), \((\pi,- 1)\), \((\frac{3\pi}{2},0)\), \((2\pi,1)\).
For \(y =-\frac{1}{2}\cos x\):

  • When \(x = 0\), \(y=-\frac{1}{2}\cos(0)=-\frac{1}{2}(1)=-\frac{1}{2}\)
  • When \(x=\frac{\pi}{2}\), \(y =-\frac{1}{2}\cos(\frac{\pi}{2})=0\)
  • When \(x=\pi\), \(y=-\frac{1}{2}\cos(\pi)=-\frac{1}{2}(-1)=\frac{1}{2}\)
  • When \(x = \frac{3\pi}{2}\), \(y=-\frac{1}{2}\cos(\frac{3\pi}{2})=0\)
  • When \(x=2\pi\), \(y=-\frac{1}{2}\cos(2\pi)=-\frac{1}{2}(1)=-\frac{1}{2}\)

Step4: Sketch the graph

Connect the key points \((0,-\frac{1}{2})\), \((\frac{\pi}{2},0)\), \((\pi,\frac{1}{2})\), \((\frac{3\pi}{2},0)\), \((2\pi,-\frac{1}{2})\) with a smooth curve. The graph has a period of \(2\pi\) (same as \(y = \cos x\) since there is no horizontal - scaling, \(B = 1\) in \(y=A\cos(Bx)\)), amplitude \(\frac{1}{2}\), and is reflected about the \(x -\)axis.

Answer:

The graph of \(y =-\frac{1}{2}\cos x\) has amplitude \(\frac{1}{2}\), is a reflection of \(y = \cos x\) about the \(x -\)axis, and has key points \((0,-\frac{1}{2})\), \((\frac{\pi}{2},0)\), \((\pi,\frac{1}{2})\), \((\frac{3\pi}{2},0)\), \((2\pi,-\frac{1}{2})\) which are connected with a smooth cosine - like curve with period \(2\pi\).