the coordinate of statianary point of the given function y=x^2_4x
0
0
Be the first one react
1 Answer
You must login to add an answer.
REGISTER! Share and grow knowledge of the world! We want to connect educated people with those who need it, to bring them together and to be able to share their knowledge with everyone. Join the Questions & Answers here.
Please sign in to your account!
Lost your password? Please enter your email address. You will receive a link and will create a new password via email.
Please briefly explain why you feel this question should be reported.
Please briefly explain why you feel this answer should be reported.
Please briefly explain why you feel this user should be reported.
You must login to add an answer.
To find the coordinates of stationary points of the function y = x^2 – 4x, we need to first find the derivative of the function and then set it equal to zero to find the critical points. The critical points correspond to the stationary points of the function.
y = x^2 – 4x y’ = 2x – 4
Setting y’ equal to zero, we get:
2x – 4 = 0
Solving for x, we get:
2x = 4
x = 2
Therefore, the critical point (and the stationary point) of the function is at x = 2.
To find the corresponding y-coordinate, we substitute x = 2 back into the original function:
y = x^2 – 4x y = 2^2 – 4(2) y = 4 – 8 y = -4
Therefore, the coordinates of the stationary point of the function y = x^2 – 4x are (2, -4).