Substitute $f\left(1\right)f\backslash left(1\backslash right)$ and $f(-1)f\backslash left(-1\backslash right)$ into the general equation of a straight line.

$\begin{array}{c}1)7=m\cdot 1+t\\ 2)3=m\cdot (-1)+t\end{array}\backslash def\backslash arraystretch\{1.25\}\; \backslash begin\{array\}\{l\}1)\backslash ;7=m\backslash cdot\; 1+t\backslash \backslash 2)\backslash ;3=m\backslash cdot\; (-1)+t\backslash end\{array\}$

Solve the system of equations.

$1)+2):10=2t\mid :21)\backslash ;+\backslash ;2):\backslash ;10=2t\; \backslash qquad\; \backslash left|:2\backslash right.$

$\Rightarrow t=5\backslash qquad\backslash Rightarrow\{t\}=5$

Plug $tt$ into $1)1)$

$7=m+5\mid -57=m+5\; \backslash qquad\; \backslash left|-5\backslash right.$

$m=2m=2$

Substitute $mm$ and $tt$ into the general equation of a straight line.

$f\left(x\right)=2x+5f\backslash left(x\backslash right)=2x+5$

$f\left(x\right)=m\cdot x+tf\backslash left(x\backslash right)=m\backslash cdot\; x+t$

$tt$ is the $yy$-axis intercept, i.e. the value at $x=0x=0$.

$t=f(0)=at=f(0)=a$

Substitute this and $f(a)=0f(a)=0$ into the general line equation.

$0=m\cdot a+a0=m\backslash cdot\; a\; +a$

Solve for $mm$.

$m={\displaystyle \frac{-a}{a}}=-1m=\backslash dfrac\{-a\}\{a\}=-1$

Plug $m=-1m\; =\; -1$ and $t=at\; =\; a$ into the general equation of the function.

$f(x)=-x+af(x)=-x+a$

$f\left(x\right)=m\cdot x+tf\backslash left(x\backslash right)=m\backslash cdot\; x+t$

Substitute $f\left(a\right)f\backslash left(a\backslash right)$ and $f\left(2a\right)f\backslash left(2a\backslash right)$ into the general equation of a straight line.

Solve the system of equations.

$2\cdot \text{I}-\text{II}:2-(-1)=2t-t2\backslash cdot\; \backslash text\{I\}\backslash ,-\backslash ,\backslash text\{II\}:\backslash \backslash \; \backslash ;2-(-1)=2t-t$

$\Rightarrow t=3\backslash Rightarrow\; t=3$

Plug $tt$ into $\text{I}\backslash text\{I\}$.

$1=m\cdot a+31=m\backslash cdot\; a+3$

Solve the equation for $mm$.

$m=(1-3)\cdot {\displaystyle \frac{1}{a}}=-{\displaystyle \frac{2}{a}}m=(1-3)\backslash cdot\backslash dfrac1a=-\backslash dfrac\{2\}\{a\}$

Substitute $m=-{\displaystyle \frac{2}{a}}m\; =\; -\backslash dfrac2a$ and $t=3t\; =\; 3$ into the general line equation.

$f(x)=-{\displaystyle \frac{2}{a}}\cdot x+3f(x)=-\backslash dfrac\{2\}\{a\}\backslash cdot\; x+3$