Question

Please Helppp Cacl AB Determining Slopes.

Answer 1

The zeros of the function are those values of $x$ that make $y=0$. So we solve

$0=\dfrac{1+50\sin x}{x^2+3}$

The denominator will always be positive, so we can multiply both sides of the equation by it to get

$0=1+50\sin x\implies \sin x=-\dfrac1{50}$
$\implies x=\arcsin\left(-\dfrac1{50}\right)+2n\pi=-\arcsin\dfrac1{50}+2n\pi$

where $n$ is any integer. If we take $n=\pm1$ we should get the two solutions immediately adjacent to the one near $x=0$ that still lie in the interval $-5\le x\le5$. So the other two zeros are $x=-\arcsin\dfrac1{50}\pm\pi$.

The tangent line to the curve at any $x$ is determined by the value of the derivative of the function at that value of $x$. So first compute the derivative:

$y=\dfrac{1+50\sin x}{x^2+3}\implies y'=\dfrac{50\cos x(x^2+3)-2x(1+50\sin x)}{(x^2+3)^2}=\dfrac{(50x^2+150)\cos x-100x\sin x-2x}{(x^2+3)^2}$

Now just plug in the values of $x$ determined above. It's helpful to note

$\cos\left(\arcsin\dfrac1{50}\right)=\dfrac{7\sqrt{51}}{50}$
$\cos(\theta+2n\pi)=\cos\theta$
$\sin(\theta+2n\pi)=\sin\theta$