Question

What is the equation, slope, x intercept, starting point and f(0)

Answer 1

as far as starting point for the line, hmm well, the issue is that the line in this case has arrowhead on each side, meaning it keeps on going to infinity both ways, so it doesn't quite have a starting point per se.

now, to get the equation of any straight line, we simply need two points off of it, let's use those two in the picture below.

`(\stackrel{x_1}{-6}~,~\stackrel{y_1}{-6})\qquad (\stackrel{x_2}{4}~,~\stackrel{y_2}{4}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{\textit{\large rise}} {\stackrel{y_2}{4}-\stackrel{y1}{(-6)}}}{\underset{\textit{\large run}} {\underset{x_2}{4}-\underset{x_1}{(-6)}}} \implies \cfrac{4 +6}{4 +6} \implies \cfrac{ 10 }{ 10 } \implies 1`

`\begin{array}ll \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{(-6)}=\stackrel{m}{1}(x-\stackrel{x_1}{(-6)}) \implies y +6 = 1 ( x +6) \\\\\\ y+6=x+6\implies {\Large \begin{array}{llll} y=x \end{array}}`