Posted by confusedtwo questions please help on Saturday, December 1, 2012 at 9:22am.
1. find the equation of the line with the given properties. Express the equation in general form or slopeintercept 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.
x^3+x^2+64x+64<0
please show all work

college algebra  Reiny, Saturday, December 1, 2012 at 10:39am
in your 3x_y= 27, since the _ is the "shift" of the  sign, I will assume you meant
3xy= 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 xaxis
let's find the xintercepts 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 xaxis for x > 1 , and below for x < 1
so solution:
x < 1
easy to show on a number line
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