How do you know for y=(x-2)(x+3)(x-4) if the line starts in the bottom left corner or top left corner

How do you know which way to draw the graph like does it start as a maximum so does the line start in the bottom left corner or a minimum which starts at the top left corner. Like how do you know without using a calculator. You have to draw the graph first
I'm so confused please explain thanks!

a) y=(x-2)(x+3)(x-4)
b) y=(x+2)(3-x)(x+4)

When you expand the brackets if the coefficient in front of x^3 is + the first hump reaches to the sky if it's negative the first hump reaches down to the ground (opposite of quadratics)

a) If you made y = 0 and expanded the brackets the x^3 would be positive meaning the first hump reaches up to the sky like a mountain you know your first intercept on the line is -3 so the line starts diagonally below that (south west) and the hump is between -3 and 2

b) One of the x's is negative and what's 2 positives times a negative? a negative so the first hump reaches down to the ground like a valley the first point is negative 4 so the line starts north west above -4

a)

Coefficient of x^3 is 1, so limit of y as x approaches -infinity = -infinity

Graph starts at bottom left

b)

y=(x+2) (3-x) (x+4)
y=(x+2) (-x+3) (x+4)

Coefficient of x^3 is -1, so limit of y as x approaches -infinity = +infinity

Graph starts at top left