Slope Fields Question 2

The slope field for a certain differential equation is shown above. Which of the following could be a solution to the differential equation with the intial condition $y(0)=1$ ?

y=\cos x

y=1-x^2

y=\sqrt{1-x^2}

y=\dfrac{1}{1+x^2}

Approach

We can rule out the first two equations, since the graph does not correspond to the shape of the cosine or quadratic.

Note that as $x$ goes to the outer bounds, the slope approaches $0$ . There appears to be a horizontal asymptote.

For $\dfrac{1}{1+x^2}$ , the horizontal asymptote is $y=0$ since the denominator approaches infinity as $x$ approaches larger values.