First, we have to find x- and y-intercepts of given equation.
1. y-intercept is a point (0, y) so substitute x = 0 and calculate:
[tex]y=x^3-3x^2+x-3\\\\y= 0^3-3\cdot0^2+0-3\\\\y=0-0+0-3\\\\\boxed{y=-3}[/tex]
y-intercept = (0, -3)
2. x-intercept is a point (or points) (x, 0), so when we substitute y = 0 there will be:
[tex]0= x^3-3x^2+x-3\\\\0=x^2(x-3)+(x-3)\\\\0=(x^2+1)(x-3)\\\\\\x^2+1=0\qquad\vee\qquad x-3=0\\\\x^2+1\text{ always}\ \textgreater \ 0\qquad\vee\qquad x=3\\\\\boxed{x=3}[/tex]
x-intercept = (3, 0)
Now, when we have x- and y-intercepts, we can write an equation of a line (graph of the linear function) that passes through this two points:
[tex]A=(3,0)\qquad\qquad B=(0,-3)\\\\\\
y-y_A=\dfrac{y_B-y_A}{x_B-x_A}(x-x_A)\\\\\\y-0=\dfrac{-3-0}{0-3}(x-3)\\\\\\y=\dfrac{-3}{-3}(x-3)\\\\\\y=1(x-3)\\\\\\\boxed{y=x-3}[/tex]
Our linear equation is: y = x - 3
