This question already has an answer here:
Set : $\displaystyle g(x)= (mx)^2+\int_0^x f(t)\ \mathrm{d}t$, $g$ is continuous since it is the sum of two continuous
(and also differentiable) functions.
We have : $g(0)=0$, and : $\displaystyle \lim_{x\to \infty} g(x)= +\infty$, for any $m\in \mathbb{R}$ .
So the equation : $g(x)=\alpha$ has a solution for any $\alpha \geq 0$.
Here we have $\alpha =2$
Consider a function
$$F(x)=y^2+\int _{ 0 }^{ x}f(t)dt-2$$.
$F(x)$ is continuous as $f(x)$ is continuous.
Putting $y=mx$ in the equation of F(x).
$$F(x)={m^2x^2}+\int _{ 0 }^{ x}f(t)dt-2$$
Clearly $F(0)=-2$
$F(x)\rightarrow\infty$ as $x\rightarrow\infty$ so somewhere F(x) must exceed 0 and hence somewhere F(x) is indeed 0.
So y=mx surely intersects the line!
P.S:My solution is quite similar to @juantheron though :-)!