Prove that the function f(x) = x 3 + x 2 + 6x + 1 has one and only one real root.

F(0)=1. When x>0 f(x)>0 because there are no negative coefficients.
f(-1)=-5, so there is a sign change between x=-1 and 0, implying there is one root between these limits.
When x<-1, x³+x²<0 and 6x<0 and so f(x)<0. Therefore there is only one real root.

Abit confused about what you mean, but i think you might be referring to slope intercept form: y = mx + bin that case b, or the y intercept is already given.(0,-1)find the slope by using the formulay2 - y1/x2 - x1the final slope will be 4/2, equivalent to 2.now your equation should look like thisy = 4/2x -1sorry if this didn’t you out!