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Point of discontinuity

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Now Playing:Point of discontinuity– Example 1
Examples
  1. Investigating How a Point of Discontinuity Appears on a Graph
    Sketch and compare the following two functions:
    i) f(x)=2x+5f\left( x \right) = 2x + 5
    ii) g(x)=2x2+11x+15x+3g\left( x \right) = \frac{{2{x^2} + 11x + 15}}{{x + 3}}
    Point of discontinuity
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    Notes
    • "point of discontinuity" exists when the numerator and denominator have a factor in common.
    i.e. (x)=−(3x−8)(x+5)(2x−7)(x+5)(4x+9)(3x+8)(2x−7)\left( x \right) = \frac{{ - \left( {3x - 8} \right)\left( {x + 5} \right)\left( {2x - 7} \right)}}{{\left( {x + 5} \right)\left( {4x + 9} \right)\left( {3x + 8} \right)\left( {2x - 7} \right)}} ; points of discontinuity exist at x=−5x = - 5 and x=72x = \frac{7}{2} .
    • 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.