2. The zero of the linear function is 2, and the initial value is $\frac{5}{2}$. Write the equation of the line that is parallel to the graph of the given function and intersects the $x$-axis at $\frac{4}{3}$, in implicit form.
2. Considering that $n=\frac{5}{2}$ and $f(2)=0$, we substitute these into the formula for a linear function $f(x)=k \cdot x+n$. We get $k \cdot 2+\frac{5}{2}=0$. Solving for $k$, we find $k=-\frac{5}{4}$. We consider the property of parallel lines, which means they have the same slope. The equation of the line is $y=-\frac{5}{4} x+n_{1}$. We consider that the point $M\left(\frac{4}{3}, 0\right)$ lies on this line and calculate $n_{1}=\frac{5}{3}$. The equation of the line is $y=-\frac{5}{4} x+\frac{5}{3}$. We write this equation in implicit form as $15 x+12 y-20=0$ or any equivalent form.

![](https://cdn.mathpix.com/cropped/2024_06_07_6d1b766ff5bddd57ba61g-09.jpg?height=68&width=1639&top_left_y=1799&top_left_x=206)

![](https://cdn.mathpix.com/cropped/2024_06_07_6d1b766ff5bddd57ba61g-09.jpg?height=57&width=1639&top_left_y=1853&top_left_x=206)

Considering the equality of the slopes $y=-\frac{5}{4} x+n_{1} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots$

![](https://cdn.mathpix.com/cropped/2024_06_07_6d1b766ff5bddd57ba61g-09.jpg?height=60&width=1642&top_left_y=1952&top_left_x=207)

![](https://cdn.mathpix.com/cropped/2024_06_07_6d1b766ff5bddd57ba61g-09.jpg?height=68&width=1642&top_left_y=1999&top_left_x=207)

![](https://cdn.mathpix.com/cropped/2024_06_07_6d1b766ff5bddd57ba61g-09.jpg?height=55&width=1639&top_left_y=2057&top_left_x=208)