if p(x)=(x-1)(x+k) and if the line tangent to the graph of p at the point (4,p(4)) is parallel to the line
5x-y+6=0, then k=?
a. -1
b. -2
c. 1
d. 0
e. 2
parallel to y = 5 x - 6 so slope = 5
so dp/dx = 5 at x = 4
dp/dx = slope(x) = (x-1)(1) + (x+k) (1)
= x - 1 + x + k = 2x+k-1
at x = 4, dp/dx = 7+k
so
7+k = 5
k =-2
Well, well, well! Let's put on our clown shoes and tackle this math problem!
To find the slope of the tangent line to the graph of p at the point (4, p(4)), we need to find the derivative of p(x) and evaluate it at x = 4. Don't worry, I won't get too serious with the math jargon!
Let's start by finding the derivative of p(x), which is the same as finding the slope of p(x) at any point. The derivative of p(x) is given by p'(x) = d(p(x))/dx, which is equal to the sum of the derivatives of each term.
So, p'(x) = (1)(x+k) + (x-1)(1).
Now, we need to evaluate p'(x) at x = 4. Plugging in x = 4, we get:
p'(4) = (1)(4+k) + (4-1)(1) = 4 + k + 3 = 7 + k.
Now, we know that the tangent line to the graph of p at the point (4, p(4)) is parallel to the line 5x - y + 6 = 0. To determine parallel lines, we compare their slopes.
The line 5x - y + 6 = 0 is in the form y = mx + b, where m is the slope of the line. So, the slope of this line is 5.
We want the slope of the tangent line to be parallel to the slope of the given line, which means p'(4) should also be equal to 5.
Setting p'(4) = 7 + k equal to 5, we get:
7 + k = 5.
Subtracting 7 from both sides, we have:
k = 5 - 7.
Simplifying further, we find:
k = -2.
So, clowning around and making some funny calculations, k is equal to -2.
Therefore, the correct answer is (b) -2.
To find the value of k, we need to determine the derivative of the function p(x) and use it to find the slope of the tangent line at (4, p(4)).
First, let's find the derivative of p(x):
p(x) = (x-1)(x+k)
Using the product rule, the derivative of p(x) is:
p'(x) = (1)(x+k) + (x-1)(1)
Simplifying, we have:
p'(x) = x + k + x - 1
p'(x) = 2x + k - 1
To find the slope of the tangent line, we substitute x = 4 into the derivative:
p'(4) = 2(4) + k - 1
p'(4) = 8 + k - 1
p'(4) = 7 + k
The slope of the line 5x - y + 6 = 0 is 5. Since the line tangent to the graph of p at (4, p(4)) is parallel to this line, their slopes must be equal.
Therefore, we have:
7 + k = 5
Solving for k:
k = 5 - 7
k = -2
So, the value of k is -2. Therefore, the correct answer is (b) -2.
To find the value of k, we need to determine the slope of the line tangent to the graph of p(x) at the point (4, p(4)), and then compare it to the slope of the given line.
Step 1: Find the derivative of p(x)
To find the slope of the tangent line, we need to take the derivative of p(x). The function p(x) is given by p(x) = (x-1)(x+k), so let's expand and simplify it:
p(x) = x^2 + (k-1)x - k
Step 2: Find the derivative of p(x)
To find the derivative, we differentiate p(x) with respect to x:
p'(x) = 2x + (k-1)
Step 3: Find the slope of the tangent line
Evaluate the derivative at x = 4 to find the slope of the tangent line:
p'(4) = 2(4) + (k-1) = 8 + (k-1) = 7 + k
Step 4: Compare the slope of the tangent line with the given line
The given line has the form 5x - y + 6 = 0, where the coefficient of x is the slope. The given line is parallel to the tangent line, so their slopes must be equal.
Comparing the slopes, we have:
Slope of the line tangent to p(x): 7 + k
Slope of the given line: 5
Therefore, we can set up the equation:
7 + k = 5
Solving the equation:
k = 5 - 7
k = -2
So, the value of k is -2, which corresponds to option b.
Therefore, the correct answer is b. -2.
parallel to y = 5 x - 6 so slope = 5
so dp/dx = 5 at x = 4
dp/dx = slope(x) = (x-1)(1) + (x+k) (1)
= x - 1 + 1 + k = x+k
at x = 4, dp/dx = 4+k
so
4+k = 5
k = 1