What’s the graph function f(x) =4(1.5)^x
1 answer:
Answer:
See below.
Step-by-step explanation:
Step 1. Find any zeroes.
(a) <em>y-intercept </em>
y = 4(1.5)^x Let x = 0
= 4(1.5)⁰
= 4 × 1
= 4
We have a <em>point at (0,4)</em>.
(b) <em>x-intercept </em>
y = 4(1.5)^x Let y = 0
0 = 4(1.5)^x Divide each side by 4
0 = (1.5)^x Take the log of each side
log0 = xlog1.5
log(0) is undefined. There is <em>no x-intercept. </em>
Step 2. Identify any asymptotes
y<em> </em>= 4(1.5)^x
y can never be negative, so the <em>x-axis is an asymptot</em>e.
Step 3. Plug in and plot a few points
Here's a table of a few points.
<u> x y </u>
-6 0.4
-4 0.8
-2 1.8
-1 2.7
0 4.0
1 6.0
2 9.0
4 20.3
6 45.6
Step 4. Check the end behaviour.
y = 4(1.5)^x
As x ⟶ ∞, y ⟶ ∞.
As x ⟶ -∞, y ⟶ 0 .
Step 5. Draw a smooth line through the points.
Your graph should look something like the one below.
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Answer: Choice C
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Explanation:
The 1/2 out front handles the vertical compression by 1/2. For example, if two points are vertically spaced by 10 units, then their new vertical distance is now (1/2)*10 = 5 units.
The x+3 in the exponent means "shift 3 units to the left". Effectively what's going on is that the old input x is now x+3, ie 3 units larger than before. This shifts the xy axis itself 3 units to the right. If we held the curve fixed in place while the xy axis moved like this, then it gives the illusion the curve is moving 3 units to the left.
Then finally the -2 at the very end shifts the curve down 2 units. This is because whatever the y coordinate is, we subtract 2 from it to do this vertical shifting. For example, the point (0,62.5) shifts 2 units down to (0,60.5)