Find k such that the line [y=5x-4] is tangent to the function : x^2-kx
Where do the graphs meet?
x^2-kx = 5x-4
x^2 - (k+5)x + 4 = 0
We want there to be a single root, so the discriminant must be zero
(k+5)^2 - 16 = 0
k = -5±4 = -9,-1
check:
f(x) = x^2 - kx
f'(x) = 2x-k
k = -9
x^2+9x = 5x-4
x = -2
slope at (-2,-14) = 5
k = -1
x^2+x = 5x-4
x = 2
slope at (2,2) = 5
f(x) = x^2 - kx
f ' (x) = 2x - k
the slope of the line y = 5x-4 is 5
so 2x-k = 5
k = 2x-5
also we want x^2 - kx = 5x-4
then: x^2 - (2x-5)(x) = 5x-4
x^2 - 2x^2 + 5x = 5x - 4
-x^2 = -4
x = ±2
then k = 4-5 = -1
or
k = -4-5 = -9
check:
case1 , if k = -1
then: x^2 + x = 5x-4
x^2 - 4x + 4 = 0
(x-2)^2 = 0
x-2 = 0
x = 2 ----> only one solution ---> only one point of contac
--> must be tangent when x=2
when k = -9
x^2 + 9x = 5x-4
x^2 + 4x + 4 = 0
(x+2)^2 = 0
x+2 = 0
x = -2 , ----> only one solution ---> only one point of contac
--> must be tangent when x=-2
Why did the line go to see the function? Because it wanted to be a tangent! Alright, let's find k to make that happen.
To find the value of k for which the line is tangent to the function, we need to equate the slopes of the line and the function. The slope of the line y = 5x - 4 is 5.
To find the slope of the function x^2 - kx, we differentiate it with respect to x. The derivative is 2x - k. Since the line and the function are tangent, their slopes must be equal. Therefore, we equate 5 to 2x - k:
5 = 2x - k
To proceed further and solve for k, we need more information. Is there anything else you can provide?
To find the value of k such that the line y = 5x - 4 is tangent to the function f(x) = x^2 - kx, we need to determine the coordinates of the point where the line and the function intersect.
First, we set the equation of the line equal to the equation of the function:
5x - 4 = x^2 - kx
Rearranging the equation, we get:
x^2 - (k + 5)x + 4 = 0
For the line to be tangent to the function, there should be exactly one solution for this equation. This means the quadratic equation should have a discriminant of zero, since a discriminant of zero implies that the quadratic equation has one real root.
The discriminant (D) of a quadratic equation ax^2 + bx + c = 0 is given by the formula:
D = b^2 - 4ac
In our equation, a = 1, b = -(k + 5), and c = 4. Substituting these values into the discriminant formula, we have:
D = (-(k + 5))^2 - 4(1)(4)
= (k + 5)^2 - 16
For the line to be tangent to the function, the discriminant should equal zero. So we set D = 0 and solve for k:
(k + 5)^2 - 16 = 0
Expanding the square and simplifying, we get:
k^2 + 10k + 25 - 16 = 0
k^2 + 10k + 9 = 0
Now we can factor or use the quadratic formula to solve for k. Factoring the equation, we have:
(k + 1)(k + 9) = 0
Setting each factor equal to zero, we get:
k + 1 = 0 --> k = -1
k + 9 = 0 --> k = -9
Therefore, there are two possible values of k that would make the line y = 5x - 4 tangent to the function f(x) = x^2 - kx, which are k = -1 and k = -9.