Question

1. A local min of y=x^3+2x^2-1 is?
2.The y-intercept of y=x^3+2x^2-1 is?
3.The y-intercept of y=x^2+4x-2 is?

Answer 1

you can factor it to find the x-intercepts.
\(\large\color{black}{ y=x^3+2x^2-1 }\)
\(\large\color{black}{ y=x^3+x^2+x^2-1 }\)
\(\large\color{black}{ y=x^2(x+1)+x^2-1 }\)
\(\large\color{black}{ y=x^2(x+1)+(x-1)(x+1) }\)
\(\large\color{blue}{ y=(x^2+x-1)(x+1) }\)
and then: go by a zero product property,
Either \(\large\color{blue}{ x^2+x-1=0 }\) or \(\large\color{blue}{ x+1=0 }\)

Answer 2

for y-intercept, set x equal zero, and all terms will become a zero, besides the constant

Answer 3

For local minimum, I don't really recall any algebraic way to do it besides graphing, but I will look that up.

Answer 4

Oh, I am thinking it would be one of the roots, after removing any vertical shift. Well, in this case it can work for the least part.
\(\large\color{black}{ y=x^3+2x^2\color{red}{-1 } }\) is shifted 1 unit down (the shift is labelled in red).

Answer 5

okay, so finding the roots we get:
\(\large\color{black}{ y=x^3+2x^2~~~~~\Rightarrow~~~~~y=x^2(x+2) ~~~~~\Rightarrow~~~~~x=~~0,~~-2 }\) And so,

Will check which is smaller

Answer 6

oh no, not check which is smaller, but a turning point, is what we need.

Answer 7

1.find dy/dx and then critical points
2.find \[\frac{ d^2y }{ dx^2 }\]

and see if it>0 at critical points for local minima.