MHB Drawing Contour Maps: Level Curves of $f(x,y)=(y-2x)^2$

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To draw a contour map of the function f(x,y)=(y-2x)^2, start by setting k equal to various positive values, which represent the level curves. For each value of k, solve the equation k=(y-2x)^2 to find the corresponding curves. This results in two equations for y: y=2x±√k, which can be plotted for different k values. The contour map will show parabolic curves that open upwards, illustrating the relationship between x and y for the given function. By plotting these curves, you can visualize how the function behaves across the xy-plane.
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draw a contour map of the function showing several level curves $f(x,y)=(y-2x)^2$

how do i do this. i know i have $k=(y-2x)^2$ but how do i use it?

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Use arbitrary positive values for $k$ and draw the resultant curve for each value.
 

Thread 'Proving that convexity implies second order derivative being positive'

There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...
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