For y=k enx+a, with non-zero integers k,n, the exponential part is positive. Therefore the curve stays above y=a when k>0 and below it when k<0. Its full-domain range excludes a. If n>0, enx approaches zero as x becomes very negative; if n<0, it approaches zero as x becomes very positive. This gives the horizontal asymptote y=a. Method: find the domain and asymptote; set x=0 for the vertical intercept k+a; solve enx=- ak for a horizontal intercept only when the right side is positive; use the signs of k,n for direction; sketch. For y=2 e-x-1, the curve decreases, has range y>-1, and crosses the axes at (0,1) and (ln2,0). Exam check: a negative multiplier reflects the curve vertically. A restricted domain can remove an intercept or the end approaching the original asymptote. A constant case k=0 or n=0 must be handled directly.

For \(y=k\mathrm e^{nx}+a\), with non-zero integers \(k,n\), the exponential part is positive. Therefore the curve stays above \(y=a\) when \(k>0\) and below it when \(k<0\). Its full-domain range excludes \(a\).

If \(n>0\), \(\mathrm e^{nx}\) approaches zero as \(x\) becomes very negative; if \(n<0\), it approaches zero as \(x\) becomes very positive. This gives the horizontal asymptote \(y=a\).

Method: find the domain and asymptote; set \(x=0\) for the vertical intercept \(k+a\); solve \(\mathrm e^{nx}=-\dfrac ak\) for a horizontal intercept only when the right side is positive; use the signs of \(k,n\) for direction; sketch.

For \(y=2\mathrm e^{-x}-1\), the curve decreases, has range \(y>-1\), and crosses the axes at \((0,1)\) and \((\ln2,0)\).

Exam check: a negative multiplier reflects the curve vertically. A restricted domain can remove an intercept or the end approaching the original asymptote. A constant case \(k=0\) or \(n=0\) must be handled directly.

Exponential graph signs and asymptotes

Exponential graph signs and asymptotes