Investigating How a Point of Discontinuity Appears on a Graph
Sketch and compare the following two functions:
i) f(x)=2x+5
ii) g(x)=x+32x2+11x+15
Sketching Rational Functions Incorporating Asymptotes and Points
of Discontinuity
Sketch the rational function:
f(x)=2x3−11x2+19x−102x2−7x+5
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• "point of discontinuity" exists when the numerator and denominator have a factor in common.
i.e. (x)=(x+5)(4x+9)(3x+8)(2x−7)−(3x−8)(x+5)(2x−7) ; points of discontinuity exist at x=−5 and x=27 .
• To determine the coordinates of the point of discontinuity:
1) Factor both the numerator and denominator.
2) Simplify the rational expression by cancelling the common factors.
3) Substitute the non-permissible values of x into the simplified rational expression to obtain the corresponding values for the y-coordinate.