graph this function: y = 2^x + 2?

Answer 1

If you're not yet into calculus, choose some values for x and solve for y, then plot these points and connect them by a smooth line.

If you are in a calculus based area, the best way to graph is to find our domain and range, x-intercepts, y-intercept, where our graph is increasing and decreasing, local extrema, concativity, and asymptotes. This will give us a general overview of what our function looks like.

Our domain is all real numbers, since there is not restriction on what we can raise a constant to. Our range can only approach 2, since two raised to anything can only get infinitely small but never zero. But since we can get as big as we want, we know our range is in the open interval (2, ∞).

Our x-intercept is the x that gives us a y of 0, so we solve:

0 = 2^x + 2
-2 = 2^x

We know that a positive number raised to any power is still a positive number, so we have no solutions that will make 2^2 = -2 and therefore we have no x-intercepts.

Our y-intercept is the y value we obtain at x = 0, so we have:

f(0) = 2^0 + 2
= 1 + 2
= 3

We have a y-intercept at P(0, 3)

To find where our slope is increasing or decreasing, we need to derive our function and find our critical points.

f(x) = 2^x + 2
f'(x) = 2^x * Ln2

We can tell right away that our derivate is always positive, so our function is always increasing. We have no minimum or maximum value, because our function values just get bigger and bigger on and on to the right, and smaller the other way. Any value we choose, there is always another bigger and smaller.

To find concativity, we need our second derivative.

f(x) = 2^x + 2
f'(x) = 2^x * Ln2
f''(x) = Ln2 * 2^x * Ln2 = 2Ln2 * 2^x

This function is also always positive, meaning our slopes are always getting larger as we move to the right, making this graph concave upward.

Finally, we want to know if we have any asymptotes as our function approaches +∞ and - ∞

Lim [x --> +∞] 2^x + 2 = +∞

The above should be fairly obvious, because as x gets larger and larger, so will our answer.

Lim [x --> -∞] 2^x + 2
= Lim[x --> +∞] 1/(2^x) + 2 = 2

As x becomes negative, 2^x is the same thing as 1/(2^|x|), and we already saw that as x grows larger and larger so will 2^x. 1 divided by this number which approaches infinity will approach 0. So we are left with a horizontal asymptote at y=2

If you think about it, our value will always be some amount larger than 2, so we know our function is above this asymptote, which we could also deduce from the fact that our range was (2, +∞).

When graphing this, start out with your y-intercept. Draw your asymptote as a guide. Then draw your function so it looks like it is always growing from left to right. Towards the left, it should get as close to your asymptote as you can make it look without actually touching, and towards the right you want to approach the highest point on your graph. This will give you an idea of how steep you need to make your curve. Mark your y-axis to show where the graph is at y=2 so we know what our asymptote is at, and show your intercept of y=3. You can leave your x-axis without value labels. Since our x axis is unscaled, we don't have to worry about how our curve increases, as long as it shows the general trend of increasing concave upward towards infinity.

--charlie

Answer 2

This would be similar to the e^x / a^x exponential function. But this will be higher from the X axis by 2. Because at x = 0 , y = 3, while in the case of , y = a^x , ( 2^x) at x = 0, y = 1.

Answer 3

You can solve the equation
( y = 2 ^ x + 2 )
manually by putting different value of X ranging from ' - ' to ' + '.But you asked that how to draw the graph of this equation.There are different Mathematical tools for drawing the graph of any equation.Every tool has it's own format such as Matlab,Maple,Mathcad etc.The tool that I've used is Maple 7 and the format how to draw it as follow.

> with(plots):
>with(plottools):
>plot(2^x+2,x=-10..10,y=-10..10,title=" y = 2^x+2 ");


The values of X and Y can be on your own choice, I've just take the temporary values.

NOTE:

If do you have any problem than you can also contact with me on my E mail address [e-mail]

Answer 4

It's the same concept as graphing any equation. Choose numbers for x and solve for y.
When x = -1, y = 2.5
When x = 0, y = 3
When x = 1, y = 4
When x = 2, y = 6
When x = 3, y = 10
Plot those points and connect the dots.

Answer 5

f (x) = x² + 2x - 3 f `(x) = 2x + 2 = 0 for turning aspect x = - a million for turning aspect f "(x) = 2 (provides a minimum turning oint) f (-a million) = a million - 2 - 3 = - 4 as a outcome (- a million , - 4) is a minimum turning aspect f (x) = (x + 3)(x - a million) Cuts x axis at x = - 3 , x = a million Cuts y axis at (0,-3) Now have factors:- (- a million , - 4) , (-3 , 0) , (a million , 0) , (0 , 3) so curve may nicely be drawn thro` those factors

Answer 6

this function is always positive the derivative is Ln 2 *2^x

for x=-infinite asymptote at y=2 for x=0 y=3

for x=+infinie , plus infinite