A discriminant reveals a forbidden output interval

The curve CC has equation y=2x2+43(x+λ+1)y=\frac{2x^2+4}{3(x+\lambda+1)}. It is given that CC has a vertical asymptote x=0x=0.

(i) Determine the value of λ\lambda. mark

(ii) Find the equation of the other asymptote of CC. marks

(iii) Prove, using an algebraic method, that CC cannot lie between −4a3<y<4a3-\frac{4\sqrt a}{3}<y<\frac{4\sqrt a}{3}, a<0a<0, where aa is to be determined. marks

(iv) When λ=0\lambda=0, the graph of y=2x2+43(x+λ+1)+ky=\frac{2x^2+4}{3(x+\lambda+1)}+k cuts the xx-axis at 2 distinct points where kk is a positive integer.

(a) Sketch the graph of y=2x2+43(x+λ+1)+ky=\frac{2x^2+4}{3(x+\lambda+1)}+k when λ=0\lambda=0, indicating clearly all asymptotes and the coordinates of the axial intercepts. marks

(b) Deduce the least possible integer value of kk. mark

[2009/NYJC/Prelim/I/7]

Total marks: 11

Solution

(i)

Asymptotes

The curve CC has equation y=2x2+43(x+λ+1)y=\dfrac{2x^2+4}{3(x+\lambda+1)}. It is given that CC has a vertical asymptote x=0x=0. Determine the value of λ\lambda.

Hint

  1. A vertical asymptote is where the denominator vanishes.

Answer · (i)

λ=−1\lambda=-1.