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July 29, 2014

July 29, 2014

Posted by **confused--two questions please help** on Saturday, December 1, 2012 at 9:22am.

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 10:39amin 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

so solution:

x < -1

easy to show on a number line

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