At what points on the curve y=(x^2)/(2x+5) is the tangent line horizontal?
y'=[(2x)(2x+5)-(2)(x^2)]/(2x+5)^2
=[2x^2+10x]/(2x+5)^2
I plugged 0 into x, and got the coordinates (0,0). How do I get the other pair of coordinates?
Thanks in advance.
To find the coordinates of the other point where the tangent line is horizontal, we need to set the derivative equal to 0 and solve for x.
So, setting y' = 0:
[2x^2 + 10x] / (2x + 5)^2 = 0
To solve this equation, we can set the numerator equal to 0:
2x^2 + 10x = 0
Now, we can factor out 2x from the equation:
2x(x + 5) = 0
Applying the zero product property, we set each factor equal to 0:
2x = 0 or x + 5 = 0
Solving these two equations, we find that x = 0 or x = -5.
To find the corresponding y-coordinates, we substitute these values of x back into the original equation y = (x^2) / (2x + 5):
For x = 0:
y = (0^2) / (2(0) + 5) = 0 / 5 = 0
So, the point is (0, 0)
For x = -5:
y = ((-5)^2) / (2(-5) + 5) = 25 / (-10 + 5) = 25 / (-5) = -5
So, the point is (-5, -5)
Therefore, the two points where the tangent line is horizontal on the curve y = (x^2) / (2x + 5) are (0, 0) and (-5, -5).
To find the points on the curve where the tangent line is horizontal, we need to find where the derivative of the function is equal to zero, since the slope of a horizontal line is zero.
You correctly found the derivative of the function: y' = [2x^2+10x]/(2x+5)^2.
To find where this derivative is zero, we set it equal to zero and solve for x:
[2x^2+10x]/(2x+5)^2 = 0
Since a fraction is equal to zero when the numerator is zero, we can set the numerator equal to zero:
2x^2 + 10x = 0
Factoring out 2x, we get:
2x(x + 5) = 0
Now we can set each factor equal to zero and solve for x:
2x = 0 --> x = 0
x + 5 = 0 --> x = -5
So, we have two possible values for x where the derivative is zero: x = 0 and x = -5.
To find the corresponding y-values, we can substitute these x-values into the original function y = (x^2)/(2x+5):
For x = 0:
y = (0^2)/(2(0)+5) = 0/5 = 0
So, we have the point (0, 0).
For x = -5:
y = ((-5)^2)/(2(-5)+5) = 25/(-5) = -5
So, we have the point (-5, -5).
Therefore, the two points on the curve where the tangent line is horizontal are (0, 0) and (-5, -5).