the graph of y= (ax+b)/(x-1)(x-4) has a horizontal tangent at the point (2,-1. Find a ANd B.
Please help I have ne idea where to start
Assuming you meant
y= (ax+b)/((x-1)(x-4))
y' = (ax^2+2bx-(4a+5b))/((x-1)(x-4))^2
Now, y'=0 when ax^2+2bx-(4a+5b) = 0
when x=2, that means
4a+4b-4a-5b=0
b=0
So, y = ax/((x-1)(x-4))
at x=2, y=-1, so
2a/-2 = -1
a = 1
and so
y = x/((x-1)(x-4))
Ah, finding the values of a and b can be a bit tricky, but fear not, I am here to help you with a touch of humor!
Let's go step by step, shall we? Since the graph has a horizontal tangent at the point (2, -1), it means that the derivative of the function equals zero at x = 2.
Now, let's differentiate the function y with respect to x. But hey, don't despair! I promise to make it somewhat enjoyable for you.
Using the quotient rule, we differentiate y = (ax + b) / [(x - 1)(x - 4)]. Don't worry; we'll get through this together!
Now, differentiating with respect to x, we get:
y' = [(x - 4)(a) + (ax+b)(1)] / [(x - 1)(x - 4)]^2
Simplifying the numerator and denominator, we have:
y' = [ax - 4a + ax + b] / [(x - 1)(x - 4)]^2
= [2ax - 4a + b] / [(x - 1)(x - 4)]^2
Now set y' equal to zero and substitute x = 2:
0 = [2a(2) - 4a + b] / [(2 - 1)(2 - 4)]^2
Simplifying further, we have:
0 = (4a - 4a + b) / (1)(4)^2
= b / 16
Since b = 0 (because b/16 = 0), we found our first answer!
Now, let's find a. We'll use the point (2, -1) that you provided. Substitute x = 2 and y = -1 into the original equation:
-1 = (a(2) + b) / [(2 - 1)(2 - 4)]
-1 = (2a + b) / (-2)
Cross-multiplying, we get:
-1(-2) = 2a + b
2 = 2a
Divide both sides by 2:
a = 1
So now we have found the values of a and b! Drum roll, please...
a = 1, and b = 0!
And there you have it! The values of a and b for the given function. Hope this helped lighten things up a bit while tackling the problem!
To find the values of a and b, we can use the fact that the graph has a horizontal tangent at the point (2,-1).
A horizontal tangent implies that the derivative of the function at that point is zero.
First, let's find the derivative of the function y = (ax + b) / ((x - 1)(x - 4)) with respect to x.
Using the quotient rule, the derivative of y is given by:
dy/dx = [(x - 4)(a) - (ax + b)(1)] / ((x - 1)(x - 4))^2
We can simplify this expression to:
dy/dx = [a(x - 4) - (ax + b)] / ((x - 1)(x - 4))^2
Now, let's evaluate the derivative at x = 2 and set it equal to zero:
0 = [a(2 - 4) - (a(2) + b)] / ((2 - 1)(2 - 4))^2
This simplifies to:
0 = [-2a - (2a + b)] / (-2)^2
0 = [-2a - 2a - b] / 4
Simplifying further:
0 = -4a - b
We now have a system of equations:
1) -4a - b = 0
2) -2a - b = 0
From equation 1, we can solve for b in terms of a:
b = -4a
Substituting b in equation 2, we get:
-2a - (-4a) = 0
Simplifying:
-2a + 4a = 0
2a = 0
a = 0
Now we can substitute the value of a back into the equation for b:
b = -4(0)
b = 0
Therefore, a = 0 and b = 0.
So the equation of the graph is y = 0x + 0, which simplifies to y = 0.
To find the values of a and b in the equation y=(ax+b)/(x-1)(x-4) where the graph has a horizontal tangent at the point (2,-1), we can use the following steps:
Step 1: Determine the derivative of the function y with respect to x. This will give us the slope of the tangent line at any point on the graph.
Step 2: Set the derivative equal to zero and solve for x to find the x-coordinate(s) of the point(s) where the tangent line is horizontal.
Step 3: Substitute the x-coordinate(s) found in Step 2 back into the original equation to solve for the corresponding y-coordinate(s).
Step 4: Use the given point (2,-1) to form a system of equations and solve for a and b.
Let's go through these steps one by one:
Step 1: Start by determining the derivative of y. The function can be rewritten as:
y = (ax + b)/(x-1)(x-4)
Expanding the denominator using partial fractions, we get:
y = (ax + b)/[(x-1)(x-4)]
= A/(x-1) + B/(x-4)
To find A and B, we need to equate the numerators:
ax + b = A(x-4) + B(x-1)
Expanding and simplifying:
ax + b = (A + B)x + (-4A - B)
The coefficients of x must be equal on both sides, and the constant terms must be equal as well:
a = A + B ...(1)
b = -4A - B ...(2)
Step 2: Set the derivative equal to zero and solve for x:
To find where the tangent is horizontal, we need to find the critical points. The derivative of y is obtained by differentiating both sides of equation (1):
a = A + B
Since we want the tangent to be horizontal, the derivative (a) must be zero. So we have:
0 = A + B
Step 3: Substitute x = 2 into the original equation to solve for y:
y = (ax + b)/(x-1)(x-4)
y = (2a + b)/(2-1)(2-4)
y = (2a + b)/-2
We know that y = -1, so:
-1 = (2a + b)/-2
Simplifying, we get:
2a + b = 2 ...(3)
Step 4: Use the given point (2,-1) to form a system of equations:
From equation (2), we have:
0 = A + B
Substituting A = -B into equation (3), we get:
2(-B) + b = 2
-2B + b = 2
b = 2 + 2B
Substituting b = 2 + 2B into equation (1), we have:
a = A + B
a = -B + B
a = 0
Finally, substituting a = 0 and b = 2 + 2B into equation (2), we get:
0 = -4A - B
0 = -4(0) - B
0 = -B
B = 0
Substituting B = 0 into equation (2), we have:
b = 2 + 2B
b = 2 + 2(0)
b = 2
So, the values of a and b are a = 0 and b = 2, respectively.