The function f is defined by f(x)=2x^3-x^2+kx-6, where k is a constant. In the xy-plane, the graph of f intersects the x-axis at the three points (3,0), (-1/2,0), and (q,0). What is the value of k?

take the first equaitions.

0=2*27-9+3k-6
or
-3k=54-9-6
3k=-39
k=-13

To find the value of k, we need to determine the x-coordinate of the point where the graph of f intersects the x-axis. Let's find the value of q.

The graph of f intersects the x-axis at the points (3,0) and (-1/2,0), which means that when x = 3 or x = -1/2, the function f(x) equals zero.

For x = 3:
f(3) = 2(3)^3 - (3)^2 + k(3) - 6
= 54 - 9 + 3k - 6
= 39 + 3k

Since f(3) = 0, we have:
39 + 3k = 0
3k = -39
k = -13

For x = -1/2:
f(-1/2) = 2(-1/2)^3 - (-1/2)^2 + k(-1/2) - 6
= -1/2 + 1/4 - k/2 - 6
= -13/4 - k/2

Since f(-1/2) = 0, we have:
-13/4 - k/2 = 0
-13 - 2k = 0
-2k = 13
k = -13/2

Now we have two values of k, k = -13 and k = -13/2. To determine which one is correct, we need to find the x-coordinate of the third point where the graph intersects the x-axis.

For x = q:
f(q) = 2q^3 - q^2 + kq - 6

Since f(q) = 0 (as the graph intersects the x-axis at the third point), we have:
2q^3 - q^2 + kq - 6 = 0

Now let's substitute in the two values of k we found earlier to solve for q.

For k = -13:
2q^3 - q^2 - 13q - 6 = 0

For k = -13/2:
2q^3 - q^2 - (13/2)q - 6 = 0

By solving either equation for q, we can find the value of k.