Math  /  Algebra

Question\#2: Graph y=2+2x2y=2+2^{x-2} using x=0,1,2,3,4x=0,1,2,3,4 (Count by 1 's up to 8 along the yy-axis) (5 Points)

Studdy Solution

STEP 1

What is this asking? We need to plot the graph of y=2+2x2y = 2 + 2^{x-2} for xx values from 0 to 4, using a scale of 1 on the x-axis and a scale of 1 up to 8 on the y-axis. Watch out! Don't forget the order of operations!
We need to calculate the exponent x2x-2 *first*, *then* apply 22 to that power, and *finally* add 22.

STEP 2

1. Calculate y-values
2. Plot the points

STEP 3

Let's **start** by calculating the values of yy when x=0x = 0.
We substitute x=0x = 0 into our equation: y=2+202=2+22y = 2 + 2^{0-2} = 2 + 2^{-2}.

STEP 4

Remember that a negative exponent means "one over"!
So, 222^{-2} is the same as 122=14=0.25\frac{1}{2^2} = \frac{1}{4} = 0.25.
Therefore, when x=0x = 0, y=2+0.25=2.25y = 2 + 0.25 = \textbf{2.25}.

STEP 5

Now for x=1x = 1: y=2+212=2+21=2+12=2.5y = 2 + 2^{1-2} = 2 + 2^{-1} = 2 + \frac{1}{2} = \textbf{2.5}.
Awesome!

STEP 6

Next, x=2x = 2: y=2+222=2+20y = 2 + 2^{2-2} = 2 + 2^0.
Anything to the power of zero is one, so 20=12^0 = 1.
This gives us y=2+1=3y = 2 + 1 = \textbf{3}.

STEP 7

Let's keep going!
For x=3x = 3: y=2+232=2+21=2+2=4y = 2 + 2^{3-2} = 2 + 2^1 = 2 + 2 = \textbf{4}.

STEP 8

Finally, for x=4x = 4: y=2+242=2+22=2+4=6y = 2 + 2^{4-2} = 2 + 2^2 = 2 + 4 = \textbf{6}.
We've got all our yy values!

STEP 9

Now, let's **plot** these points on our graph.
Remember, we're counting by 1's up to 8 on the y-axis.

STEP 10

Our points are: (0,2.25)(0, \textbf{2.25}), (1,2.5)(1, \textbf{2.5}), (2,3)(2, \textbf{3}), (3,4)(3, \textbf{4}), and (4,6)(4, \textbf{6}).
Mark these points on your graph.

STEP 11

Once you have all the points marked, **connect** them with a smooth curve.
This curve represents the graph of y=2+2x2y = 2 + 2^{x-2}.

STEP 12

The graph of y=2+2x2y = 2 + 2^{x-2} is plotted with the points (0,2.25)(0, 2.25), (1,2.5)(1, 2.5), (2,3)(2, 3), (3,4)(3, 4), and (4,6)(4, 6) connected by a smooth curve.

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