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.
x^3+x^2+64x+64<0
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
so solution:
x < -1
easy to show on a number line
1. To find the equation of a line perpendicular to another line, we need to find the negative reciprocal of the slope of the given line. First, let's rewrite the given equation in slope-intercept form by solving for y.
-3x + y = -27
y = 3x - 27
The slope of this line is 3. The negative reciprocal of 3 is -1/3. Now, we have the slope of the perpendicular line.
We also know that the perpendicular line contains the point (9, -1). We can use the point-slope form of a linear equation to find the equation of the line.
y - y1 = m(x - x1)
Plugging in the values, we have:
y - (-1) = -1/3(x - 9)
y + 1 = -1/3x + 3
y = -1/3x + 2
Therefore, the equation of the line perpendicular to the given line (-3x + y = -27) and passing through the point (9, -1) is y = -1/3x + 2.
2. To solve the polynomial inequality x^3 + x^2 + 64x + 64 < 0, we need to find the values of x that make the expression on the left side less than zero.
First, let's factor the polynomial if possible. Unfortunately, it doesn't have any obvious factoring. So, we'll need to use other methods.
To solve this inequality, we can use a sign chart or test intervals. Here, let's test intervals:
1. Choose a test value in the first interval, let's say -10. Plug it into the inequality:
(-10)^3 + (-10)^2 + 64(-10) + 64 < 0
-1000 + 100 + (-640) + 64 < 0
-1476 < 0
The inequality is true for this interval.
2. Choose a test value in the second interval, let's say 0. Plug it into the inequality:
0^3 + 0^2 + 64(0) + 64 < 0
64 < 0
The inequality is false for this interval.
3. Choose a test value in the third interval, let's say 10. Plug it into the inequality:
10^3 + 10^2 + 64(10) + 64 < 0
2764 < 0
The inequality is false for this interval.
Now, we can conclude that the solution set lies in the first interval (-∞, -infinity), where the inequality is true.
Therefore, the solution set in interval notation is (-∞, -infinity).