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 \(y(0)=1\) ?

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.