Equation of a line that is perpendicular to the graph of y = 2/5x-1
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So essentially, this is a cubic function where T(x) = y, and:
In order to find the vertices, aka "turning points", we need to see around what area the graph will curve quickly from bottom-left to top-right. If the value being cubed (in the parentheses) is around -8, y will be close to 0, since -8+7=-1. (-2)^3 = -8. So x+5=-2, x=-7.
At x=-6, y = (-6+5)^3+7 = (-1)^3+7
y = -1+7 = 6
At x=-5, y = 0+7 = 7
Above x=-4, the y jumps rapidly above 10, and below x=-9, the y jumps rapidly below -50
What does all that mean???
That between x = -9 and x = -4 (x = -4,-5,-6,-7,-8,-9), the temperature (T(x) or y) does not change very much. Therefore, around those temperatures is a turning point, where the phase shift occurs from solid/liquid.
In order to find the vertices, aka "turning points", we need to see around what area the graph will curve quickly from bottom-left to top-right. If the value being cubed (in the parentheses) is around -8, y will be close to 0, since -8+7=-1. (-2)^3 = -8. So x+5=-2, x=-7.
At x=-6, y = (-6+5)^3+7 = (-1)^3+7
y = -1+7 = 6
At x=-5, y = 0+7 = 7
Above x=-4, the y jumps rapidly above 10, and below x=-9, the y jumps rapidly below -50
What does all that mean???
That between x = -9 and x = -4 (x = -4,-5,-6,-7,-8,-9), the temperature (T(x) or y) does not change very much. Therefore, around those temperatures is a turning point, where the phase shift occurs from solid/liquid.