How do you sketch the graph #f(x)=2x^4-26x^2+72#?

Question

Christopher Agresta · Accepted Answer

See explanation...

#f(x) = 2x^4-26x^2+72# #color(white)(f(x)) = 2((x^2)^2-13(x^2)+36)# #color(white)(f(x)) = 2(x^2-4)(x^2-9)# #color(white)(f(x)) = 2(x-2)(x+2)(x-3)(x+3)# So the graph of this function intercepts the #x# axis at #x=+-2# and #x=+-3#. It intercepts the #y# axis at #(0, 72)#, where it has a local maximum. Note that #f(x)# is an even function, symmetric about the #y# axis. #f'(x) = 8x^3-52x = 2((2x)^2-26)x# So this quartic has local minima at: #x = +-sqrt(26)/2 ~~ +-5.1/2 = +-2.55# We find: #f(+-sqrt(26)/2) = 2(13/2)^2-26(13/2)+72 = 169/2-169+72 = -25/2# So this quartic function is a classic "W" shaped curve, symmetric about the #y# axis, passing through: #(-3, 0), (-sqrt(26)/2, -25/2), (-2, 0), (0, 72), (2, 0), (sqrt(26)/2, -25/2), (3, 0)# graph{2x^4-26x^2+72 [-10, 10, -20, 80]}

Axel Alvarez · Answer

undefined To sketch the graph of $ f(x) = 2x^4 - 26x^2 + 72 $, follow these steps:

1. Find the critical points by setting the derivative equal to zero and solving for $ x $.
2. Determine the behavior of the function around these critical points by analyzing the sign of the derivative.
3. Identify any intercepts by setting $ f(x) $ equal to zero and solving for $ x $.
4. Determine the end behavior of the function as $ x $ approaches positive and negative infinity.
5. Plot the critical points, intercepts, and end behavior, and then sketch the curve accordingly.