Respuesta :
[tex]\bf f(\stackrel{x}{0})=\stackrel{y}{7}\qquad \qquad \qquad f(\stackrel{x}{2})=\stackrel{y}{12}[/tex]
so we're really looking for the equation of a line that runs through (0,7) and (2,12),
[tex]\bf (\stackrel{x_1}{0}~,~\stackrel{y_1}{7})\qquad (\stackrel{x_2}{2}~,~\stackrel{y_2}{12}) \\\\\\ % slope = m slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{12-7}{2-0}\implies \cfrac{5}{2} \\\\\\ % point-slope intercept \stackrel{\textit{point-slope form}}{y- y_1= m(x- x_1)}\implies y-7=\cfrac{5}{2}(x-0) \\\\\\ y-7=\cfrac{5}{2}x\implies y=\cfrac{5}{2}x+7[/tex]
so we're really looking for the equation of a line that runs through (0,7) and (2,12),
[tex]\bf (\stackrel{x_1}{0}~,~\stackrel{y_1}{7})\qquad (\stackrel{x_2}{2}~,~\stackrel{y_2}{12}) \\\\\\ % slope = m slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{12-7}{2-0}\implies \cfrac{5}{2} \\\\\\ % point-slope intercept \stackrel{\textit{point-slope form}}{y- y_1= m(x- x_1)}\implies y-7=\cfrac{5}{2}(x-0) \\\\\\ y-7=\cfrac{5}{2}x\implies y=\cfrac{5}{2}x+7[/tex]
f(x)=ax+b
f(0)=a*0+b
7=0+b
b=7
f(2)=2a+7
12=2a+7
2a=12-7=5
a=2.5
f(x)=2.5x+7
f(0)=a*0+b
7=0+b
b=7
f(2)=2a+7
12=2a+7
2a=12-7=5
a=2.5
f(x)=2.5x+7
