Case 1 : y = ax^n , where n = 0
Linear graph where the value of a determines the gradient and c represents the y-intercept.
Since a^0 = 1 and a must be a real number, y = 1 + c .
Equation 1 : y=2x^0
=2(1)
=2
=-2(1)
=-2
Equation 3 : y=0x^0
=0(1)
=0
There is also a lack of a turning point, since the gradient doesn't change.
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