= Solution
For a <Matrix Lie group>, the <exponential map of a matrix Lie group> is the convergent <matrix exponential>
$$
\boxed{\operatorname{Exp}(X)=e^X=\sum_{n=0}^{\infty}\frac{X^n}{n!}.}
$$
Its domain is the <Lie algebra> $\mathfrak g=T_I G$, and $e^{tX}\in G$ for every real $t$ when $X\in\mathfrak g$. Because multiples of the same matrix commute, $e^{sX}e^{tX}=e^{(s+t)X}$ and $(e^{tX})^{-1}=e^{-tX}$. Thus take $I=\mathbb R$ to obtain the image of a <one-parameter subgroup>.
Here is the precise <classification of nontrivial one-parameter subgroups>. For $X\ne0$, the smooth <group homomorphism> $\gamma_X:t\mapsto e^{tX}$ has a closed kernel $K\subset\mathbb R$. A closed subgroup of $\mathbb R$ is zero, $\tau\mathbb Z$ for some $\tau>0$, or all of $\mathbb R$. Indeed, if positive elements have infimum zero, their integer multiples approximate every real number, so closedness makes the subgroup all of $\mathbb R$; otherwise the infimum is attained and is its least positive generator. The last possibility is excluded by $\gamma_X'(0)=X\ne0$. Give the image the quotient topology and smooth structure from $\mathbb R/K$. Its inclusion in $G$ is an injective <immersion>, since $\gamma_X'(t)=e^{tX}X$ never vanishes. Consequently it is a <Lie subgroup>, with
$$
\boxed{G_X\cong(\mathbb R,+)\quad\text{or}\quad\mathbb R/(\tau\mathbb Z)\cong S^1.}
$$
In the periodic case $[0,\tau)$ is also a parameter interval covering the image, with endpoints understood modulo $\tau$.
Two qualifications matter. If $X=0$, the image is the trivial group, a third possibility omitted by the wording “two”. Also, a general <one-parameter subgroup> need not be closed or embedded: an irrational winding $t\mapsto\operatorname{diag}(R(t),R(\sqrt2t))$ in a two-torus is an example, where $R(t)$ is a planar rotation. The classification uses the intrinsic immersed <Lie subgroup> structure, rather than assuming that the matrix subspace topology is the topology of $\mathbb R$.
For the real <special linear group>, differentiating $\det(I+tX)=1+t\operatorname{tr}X+O(t^2)$ at the identity shows that its Lie algebra consists of traceless matrices. Conversely $\det e^{tX}=e^{t\operatorname{tr}X}=1$ for a traceless matrix. Hence
$$
\boxed{\mathfrak{sl}_2(\mathbb R)=\left\{\begin{pmatrix}a&b\\c&-a\end{pmatrix}:a,b,c\in\mathbb R\right\}.}
$$
Choose the standard basis of the real <sl2 Lie algebra>,
$$
L_0=\begin{pmatrix}1&0\\0&-1\end{pmatrix},\qquad
L_+=\begin{pmatrix}0&1\\0&0\end{pmatrix},\qquad
L_-=\begin{pmatrix}0&0\\1&0\end{pmatrix}.
$$
Their squares have the required values. Direct matrix multiplication gives the <Lie brackets>
$$
\boxed{[L_0,L_+]=2L_+,\qquad[L_0,L_-]=-2L_-,\qquad[L_+,L_-]=L_0.}
$$
Writing $[L_a,L_b]=f_{ab}{}^cL_c$, these specify all the nonzero <Lie algebra structure constants>: $f_{0+}{}^+=2$, $f_{0-}{}^-=-2$, $f_{+-}{}^0=1$, together with their negatives on reversing the lower indices.
The three requested exponentials are
$$
\begin{aligned}
e^{tL_0}&=\begin{pmatrix}e^t&0\\0&e^{-t}\end{pmatrix},&&t\in\mathbb R,\\
e^{tL_+}&=I_2+tL_+=\begin{pmatrix}1&t\\0&1\end{pmatrix},&&t\in\mathbb R,\\
e^{t(L_+-L_-)}&=\begin{pmatrix}\cos t&\sin t\\-\sin t&\cos t\end{pmatrix},&&t\in[0,2\pi).
\end{aligned}
$$
Thus the first two images are isomorphic to the additive real group, while the last is $SO(2)$, the <circle group>. The PDF gives $L_+-L_-$ here; the local TeX's $L_--L_-$ is a transcription error.
For a general generator, put $\Delta=\alpha_0^2+\alpha_+\alpha_-$. Direct multiplication gives $X^2=\Delta I_2$. If $\Delta>0$, the eigenvalues $\pm\sqrt\Delta$ give an unbounded exponential image. If $\Delta=0$ but $X\ne0$, it is a <nilpotent linear map> with exponential $I_2+tX$, again unbounded. If $\Delta=-\omega^2<0$, the power series instead gives
$$
e^{tX}=\cos(\omega t)I_2+\frac{\sin(\omega t)}\omega X,
$$
with least positive period $2\pi/\omega$. Its image is a continuous image of a circle and is compact. Equivalently, $X/\omega$ is a real complex structure and is similar over $\mathbb R$ to the rotation generator. Therefore the exact criterion for <compact one-parameter subgroups of SL2R> is
$$
\boxed{G_X\text{ is compact}\quad\Longleftrightarrow\quad\alpha_0^2+\alpha_+\alpha_-<0\ \text{or}\ (\alpha_0,\alpha_+,\alpha_-)=(0,0,0).}
$$
For a nonzero generator only the strict inequality remains.
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