What function describes a line with slope 4/3 and passing through the point (2,5) ?

For this case we have that by definition, the equation of the line of the slope-intersection form is given by:[tex]y = mx + b[/tex]Where:m: It's the slopeb: It is the cut-off point with the y axisAccording to the data of the statement we have to:[tex]m = \frac {4} {3}[/tex]Thus, the equation is of the form:[tex]y = \frac {4} {3} x + b[/tex]We substitute the given point [tex](x, y) :( 2,5)[/tex] and find the cut-off point:[tex]5 = \frac {4} {3} (2) + b\\5 = \frac {8} {3} + b\\5- \frac {8} {3} = b\\b = \frac {3 * 5-8} {3}\\b = \frac {15-8} {3}\\b = \frac {7} {3}[/tex]Thus, the equation is: [tex]y = \frac {4} {3} x + \frac {7} {3}[/tex]Finally, we have that the function is of the form[tex]y = f (x),[/tex] where [tex]f (x)=\frac {4} {3} x + \frac {7} {3}[/tex]ANswer:[tex]f (x) = \frac {4} {3} x + \frac {7} {3}[/tex]