28 – The Slant Asymptote No Calculator Rational Function Investigations 28 – The Slant Asymptote No Calculator

Polynomial Division Review B E H K A D G J C F I L Compute Since dividing by 2x + 3, place 2x + 3 in the first column of your Area Model

Compute B E H K A D G J C F I L

is the remainder to the division. Compute B E H K A D G J C F I L What goes in Box G? is the remainder to the division. What goes in Box H? What goes in Box I? What goes in Box J?

Compute B F J A E I N C G K M D H L

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Compute each division below: x –5 –5x 6 2 2x –10 x 4 4x 6 –1 –x –4 x –1 –x –10 2 2x –2

Compute each division below: x –2 –2x –10 1 x 1 –1 –x x 4x 0x –1 –4 –4x

Slant Asymptote If the degree of the numerator is EXACTLY one more than the degree of the denominator, the graph of the function will have a ‘slant asymptote’. The equation of the slant asymptote is the result of ‘numerator divided by denominator’ (ignoring the remainder) A function whose graph has a ‘slant asymptote’ will not have a horizontal asymptote. The graph of f(x) will have a slant asymptote: y = x – 5. The graph of f(x) will have a slant asymptote: y = x + 4.

The graph of f(x) will have a slant asymptote: y = x – 1.

horizontal intercepts [f(x) = 0] –1, 4 vertical intercepts [x = 0] vertical asymptote at x = –2 slant asymptote y = x – 5 hole: none

horizontal intercepts [f(x) = 0] –2, –1 vertical intercepts [x = 0] vertical asymptote at x = 1 slant asymptote y = x + 4 hole: none

horizontal intercepts [f(x) = 0] –4, 3 vertical intercepts [x = 0] vertical asymptote at x = –2 slant asymptote y = x – 1 hole: none

horizontal intercepts [f(x) = 0] –3, 4 vertical intercepts [x = 0] vertical asymptote at x = –1 slant asymptote y = x – 2 hole: none

horizontal intercepts [f(x) = 0] vertical intercepts [x = 0] vertical asymptote at x = 1 slant asymptote y = x + 1 hole: none

horizontal intercepts [f(x) = 0] 1 vertical intercepts [x = 0] vertical asymptote at x = –2 vertical asymptote at x = 2 slant asymptote y = x