posted by confused--two questions please help .
1. find the equation of the line with the given properties. Express the equation in general form or slope-intercept form.
perpendicular to the line -3x_y= -27; contains the point (9, -1)
the equation of the line is ?
2. solve the polynomial inequality and then graph the solution set on a real number line. Express the solution set in interval notation.
please show all work
in your -3x_y= -27, since the _ is the "shift" of the - sign, I will assume you meant
-3x-y= -27 and you have a typo. (But then why would it not have been written as 3x + y = 27 ??? , strange)
anyway, the slope of the given line is -3
so the slope of the new line must be +1/3
and the equation has to have the form
x - 3y = c, where the c has to be found
but (9,-1) is to be on this new line, so
9 - 3(-1) = c
c = 12
new line , using my assumption, is
x - 3y = 12
2. let y = x^3 + x^2 + 64x + 64 , a standard looking cubic
so x^3+x^2+64x+64<0 must be the part below the x-axis
let's find the x-intercepts of the zeros of our function
x^3+x^2+64x+64 = 0
x^2(x+1) + 64(x+1) = 0
(x+1)(x^2 + 64) = 0
one real root: x = -1 and 2 complex roots
so a quick sketch would show that the curve is above the x-axis for x > -1 , and below for x < -1
x < -1
easy to show on a number line