the Given
[tex]3y=4-\frac{1}{2}x[/tex]We are to find the gradient and the intercepts on the x and y axes
The general form of equation of a line is
[tex]y=mx+c_{}[/tex]Simplifying the given equation we have
Dividing through by 3
[tex]y=\frac{4}{3}-\frac{1}{6}x[/tex]That is
[tex]y=-\frac{1}{6}x+\frac{4}{3}[/tex]Comparing this equation with the general equation of a line we have
Gradient = -1/6
x-intercept is at y = 0
Therefore
[tex]\begin{gathered} 0=-\frac{1}{6}x+\frac{4}{3} \\ \frac{1}{6}x=\frac{4}{3} \\ 3x=24 \\ x=8 \end{gathered}[/tex]Therefore x-intercept is (8,0)
y-intercept is at x = 0
Therefore
[tex]\begin{gathered} y=-\frac{1}{6}(0)+\frac{4}{3} \\ y=\frac{4}{3} \end{gathered}[/tex]Therefore, y-intercept = (0,4/3)
