Ok, so the equation is ((x+2)^2)-4
so how does this equation have the shape of an absolute value function and not one of the quadratic?
Please explain and show steps. Would be greatly appreciated.
USING GRAPHICAL INTERPRETATION
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f(x) = x
f(x) = x graph has V-SHAPED with the vertex at (0, 0).
g(x) = x + 2
The graph of [ f(x) = x ] shift up 2 units on Y axis. It still has the same V-SHAPED with the vertex at (0, 2)
h(x) = ( x + 2 )^2 = (x + 2) (x + 2) = x^2 + 4x + 4 = x^2 + 4x + 4
g(x) = ( x + 2 ) function multiply twice. On the h(x) graph, while the x values stays the same, the [ y = g(x) ] values get larger. It sill has V-SHAPED, but NARROWER V-SHAPED than g(x) = ( x + 2 ) with the vertex at (0, 4).
For example,
g(x) = x + 2 has (0, 2), (-10, 12), and (10, 12)
V-Shaped.
h(x) = ( x + 2 )^2 has (0, 4), (-10, 144), and (10, 144)
Narrower V-Shaped.
j(x) = ( ( x + 2 )^2 ) - 4
The graph of h(x) = ( ( x + 2 )^2 ) shift down 4 units on Y axis. It still has the same V-SHAPED with the vertex at (0, 0). Shapes Shocks and Deformations I::
What kind of geometry then do we require for shape And how does one de ne a metric this is equivalent to running the shape through the linear heat equation
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screen of the imaging means and an equation expressing a plane representing the lightflux reflected on the stainless mirror 205 does not produce an
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It just looks like an absolute value function to the naked eye. There's some curve to the graph though. It's kind of a combination of an absolute value function and a quadratic function, not two linear functions as in the case of a standard absolute value function. Try to find a linear function that fits the given function. I guarantee you can't do it.
Shape Metamorphism using p -Laplacian Equation::
For example: (1) Why does the distance. transformation work? ( 2) Are there alternative methods? curves have been inserted. In this experiment, In this
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Let's evaluate the function at y = 5 and y = 12.
5 = [((abs(x) + 2)^2] - 4
9 = ((abs(x) + 2)^2
3 = abs(x) + 2
1 = abs(x)
x = +/- 1
12 = [((abs(x) + 2)^2] - 4
16 = ((abs(x) + 2)^2
4 = abs(x) + 2
2 = abs(x)
x = +/- 2
So the graph of your function contains the following points: (1,5), (-1,5), (2,12), and (-2,12) -- and obviously includes the point (0,0). If you graph two linear functions that pass through the points (+/-1,5) and (+/-2,12), then you'll be able to see that the graphs of the linear functions don't match the graph of the given function, because the graphs of the linear functions will not include the point (0,0).
Another way to see the shape of this graph would be to simply zoom out. Set your graphing calculator to display the following parameters: x=[-10,10], y=[0,100]. This way you'll be able to see the curvature better than you would using the standard parameters.
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