Find the values of k for which the line y=1-2kx does not meet the curve y=9x²-(3k+1)x+5

Let's equate the two given functions and attempt to solve for x:

y = 1 -2kx = y = 9x^2 -(3k+1)x + 5

Eliminating y, 1 -2kx = 9x^2 -(3k+1)x + 5

Rearranging terms in descending order by powers of x:

0 = 9x^2 - (3k+1)x + 2kx + 5 - 1 , or

0 = 9x^2 - kx - x + 4

This is a quadratic equation with coefficients a = 9, b = -(k+1) and c = 4.

For certain k, not yet known, solutions exist. Solutions here implies points at which the two curves intersect.

k+1 plus or minus sqrt( [-(k+1)]^2 - 4(9)(4) )

x =

2(9)

The discriminant is k^2 + 2k + 1 - 144, or k^2 + 2k - 143.

If the discriminant is > 0, there are two real, unequal roots. We don't want this, since we're interested in finding k value(s) for which there's no solution.

If the discr. is = 0, there are two real, equal roots. Again, we don't want this.

If the discr. is < 0, there are no real roots. This is the case that interests us.

So our final task is to determine the k values for which the discr. is < 0:

Determine the k value(s) for which the discriminant, k^2 + 2k - 143, is 0.

This k^2 + 2k - 143 factors as follows: (k-11)(k+13), and when set = to 0, results in k: {-13,11}.

Set up intervals on the number line: (-infinity, - 13), (-13, 11) and (11, infinity).

Choosing a test number from each interval, determine the interval or intervals on which the discriminant is negative:

Case 1: k = -15; the discriminant (k^2 + 2k - 143) is (-15)^2 + 2(-15) - 143 = +52. Reject this interval

Case 2: k = 0; the discriminant is then 0 + 0 - 143 (negative); thus, the discriminant is negative on the interval (-13,11).

Case 3: k = 20; the discriminant is positive. Reject this interval.

Summary: The curves do not intersect on the interval (-13,11).

The discriminant of the function must be negative, for no solution to exist.

to plot points   for x just start at the origin (middle) and go to the left however many times it say(if the number is a positive) if the number is a negative then go right however many times it says if it is 0 go nowhere

now keep you pencil there and go up however many times it says on y(if it is positive) if it is negative then go down however many times it says. if it says 0 go nowhere.

you have just plotted a point! now when you finish plotting all of them connect them with a ruler

plot each point as (-3, 7), (0,4), (3,1), (8,-4)   in the parenthesis on the left of the comma is x and the right is y.

he has now walked a total of what?

if i know what this is maybe i can

step-by-step explanation: