How do you graph $y=2xx3^(x-2)-1$ ?

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Answer 1 · 2017-06-15T18:40:49

See explanation

$color(blue)("Determine the x-intercept")$

Set $y=0=2(3^(x-2))-1$

$1=2(3^(x-2))$

$3^(x-2)=1/2$

Lake logs

$(x-2)ln(3)=ln(1/2)$

$x=(ln(1)-ln(2))/ln(3) +2$

$x~~1.37$ to 2 decimal places
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$color(blue)("Determine the y-intercept")$

Set $y=2(3^(0-2))-1$

$y=2/(3^2)-1 = -7/9$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$color(blue)("Determine the extremities")$

Set $y=k" "$ where $" "k=lim_(x->+oo) 2(3^(x-2))-1$

$" "k->2oo-1 = oo$

Thus for $x ->oo" we have "y->oo$

Set $y=k" "$ where $" "k=lim_(x->-oo) 2(3^(x-2))-1$

$" "k->2/oo-1 = 0-1 = -1$
This is an asymptote.

Tony B Tony B

How do you graph y=2xx3^(x-2)-1y=2xx3^(x-2)-1?