Question

consider the function $y = a\cos(bx)$, where $a > 0$ and $b > 0$. which change affects the locations of the $x$-intercepts of the graph?
change the sign of $a$.
change the sign of $b$.
decrease $b$.
decrease $a$.

Explanation:

Step1: Find x-intercepts of original function

To find x-intercepts, set $y=0$:
$$a\cos(bx) = 0$$
Since $a>0$, we solve $\cos(bx)=0$. The general solution for $\cos(\theta)=0$ is $\theta=\frac{\pi}{2}+k\pi, k\in\mathbb{Z}$, so:
$$bx = \frac{\pi}{2}+k\pi \implies x = \frac{\pi(2k+1)}{2b}, k\in\mathbb{Z}$$

Step2: Analyze effect of changing sign of a

If $a$ becomes negative, set $y=0$:
$$-a\cos(bx)=0$$
Again, $-a
eq0$, so $\cos(bx)=0$, giving the same $x=\frac{\pi(2k+1)}{2b}, k\in\mathbb{Z}$. No change to intercepts.

Step3: Analyze effect of changing sign of b

If $b$ becomes negative, let $b'=-b, b'<0$. Solve $\cos(b'x)=0$:
$$b'x = \frac{\pi}{2}+k\pi \implies x = \frac{\pi(2k+1)}{2b'} = \frac{\pi(2k+1)}{-2b} = -\frac{\pi(2k+1)}{2b}$$
Since $k$ is any integer, $-\frac{\pi(2k+1)}{2b}$ is the same set as $\frac{\pi(2k+1)}{2b}$ (substitute $k$ with $-k-1$). No change to intercept locations.

Step4: Analyze effect of decreasing b

Let $b_1 < b, b_1>0$. The new intercepts are:
$$x = \frac{\pi(2k+1)}{2b_1}$$
Since $b_1 < b$, $\frac{\pi(2k+1)}{2b_1} > \frac{\pi(2k+1)}{2b}$, so the intercept locations shift outward along the x-axis. This changes their locations.

Step5: Analyze effect of decreasing a

Similar to changing sign of $a$, decreasing $a$ (still $a>0$) gives $y=a_1\cos(bx)=0$, so $\cos(bx)=0$, same intercepts as original. No change.

Answer:

Decrease b.