Further exercises: Linear functions

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

$\begin{array}{l}1)7=m\cdot 1+t\\ 2)3=m\cdot (-1)+t\end{array}$

Solve the system of equations.

$1)+2):10=2t|:2$

$\Rightarrow t=5$

Plug $t$ into $1)$

$7=m+5|-5$

$m=2$

Substitute $m$ and $t$ into the general equation of a straight line.

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

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

$t$ is the $y$-axis intercept, i.e. the value at $x=0$.

$t=f(0)=a$

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

$0=m\cdot a+a$

Solve for $m$.

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

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

$f(x)=-x+a$

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

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

$\begin{array}{rrclcr}\text{I}& 1& =& m\cdot a& +& t\\ \text{II}& -1& =& m\cdot 2a& +& t\end{array}$

Solve the system of equations.

$\begin{array}{l}2\cdot \text{I}-\text{II}:\\ 2-(-1)=2t-t\end{array}$

$\Rightarrow t=3$

Plug $t$ into $\text{I}$.

$1=m\cdot a+3$

Solve the equation for $m$.

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

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

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