Question 907849
{{{ f(x) = mx + b}}}, where {{{m=slope}}} and {{{b=y-intercept}}}


given: the {{{x-intercept=-1/3}}} and {{{y-intercept= 1/7}}}

 we know if we set {{{x=0}}} we got {{{y-intercept= 1/7}}}

{{{ f(0) = m*0 + 1/7}}} => {{{ f(0) =  1/7}}}

so, {{{y-intercept}}} is at point ({{{ 0 }}},{{{1/7}}})

and

if we set {{{f(x)=0}}} we got {{{x-intercept= -1/3}}}

so, {{{x-intercept}}} is at point ({{{ -1/3 }}},{{{ 0 }}})


use these two points to find a slope {{{m}}}

 
*[invoke slope 0, "1/7", "-1/3", 0]

now you have your {{{f(x) = mx + b}}} and that is {{{f(x) = 1*x + 1/7}}} or just 
{{{f(x) = x + 1/7 }}}

graph:

 {{{ graph( 600, 600, -3, 3, -3, 3, x + 1/7) }}}