# Sketch the graph for the following. y = 3(2^x-4)-2

### 2 Answers | Add Yours

Sketch the graph of `y=3(2^(x -4))-2` :

The parent graph is `y=2^x` . We have performed the following transformations:

(1) MUltiplying by 3 performs a vertical stretch -- each y-value is multiplied by 3.

(2) Subtracting 4 in the parantheses translated (shifted) the graph 4 units to the right.

(3) Subtracting 2 outside the function translated the graph 2 units down.

The function `2^x` in black, `3*2^x` in blue, `3(2^(x-4))` in green, and `3(2^(x-4))-2` in red which is the answering sketch.

Follow a point -- (0,1) is on `y=2^x` ; multiplying the y-value by 3 gives the point (0,3) on `y=3*2^x` ; translating 4 units right gives the point (4,3) on `y=3(2^(x-4))` ; and finally translating 2 units down gives the point (4,1) on `y=3(2^(x-4))-2` .

You need to open the brackets and perform the possible algebraic operations, such that:

y = 3*2^x - 12 - 2 => y = 3*2^x - 14

You need to evaluate x axis intercept, hence, you need to consider y = 0, such that:

0 = 3*2^x - 14 => 3*2^x = 14 => 2^x = 14/3

Taking common logarithms both sides, yields:

ln 2^x = ln (14/3) => x = (ln (14/3))/(ln 2)

You need to evaluate y axis intercept, hence, you need to consider x = 0, such that:

y = 3*2^0 - 14 => y = -11

Hence, evaluating the points where the graph intersects x and y axis, yields (0,-11) and ((ln (14/3))/(ln 2), 0).

You have two points, hence, you may sketch the graph, but, you need to test if the curve is concave up or down. You need to perform the second derivative test, such that:

(dy)/(dx) = 3*2^x*ln 2 => (dy)/(dx) = 2^x*ln 2^3

(dy)/(dx) = 2^x*ln 8

(d^2y)/(dx^2) = 2^x*ln 2*ln 8 = 3*2^x*(ln^2 2) > 0 => the graph of the function is concave up.

Sketching the graph of the function, under the given conditions, yields: