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Math  /  Algebra

QuestionFor questions 17 -20: Graph the function and the asymptote. Label any intercept point(s) for function. State the equation for the asymptote. [7 points each]
17. y=3xy=3^{x}
18. y=−3(12)xy=-3\left(\frac{1}{2}\right)^{x} \begin{tabular}{c|c} xx & y=112)2\left.y=1 \frac{1}{2}\right)^{2} \\ \hline-2 & 4 \\ -1 & 2 \\ 0 & 1 \\ ii & 0.5 \\ 2 & 0.25 \end{tabular} \begin{tabular}{c|c} xx & −3y-3 y \\ \hline-2 & -12 \\ -1 & -6 \\ 0 & =3=3 \\ 1 & =1.5=1.5 \\ 2 & -0.7 \end{tabular}

Studdy Solution

STEP 1

What is this asking? We need to graph y=3xy = 3^x and y=−3(12)xy = -3(\frac{1}{2})^x, find their asymptotes, and label any points where the graphs cross the x-axis or y-axis. Watch out! Exponential functions have horizontal asymptotes, not vertical ones!
Also, remember that anything raised to the power of zero equals **one**, not zero!

STEP 2

1. Graph y=3xy = 3^x
2. Graph y=−3(12)xy = -3(\frac{1}{2})^x

STEP 3

Let's pick some easy values for xx and see what yy becomes.
If x=0x = 0, then y=30=1y = 3^0 = \mathbf{1}.
If x=1x = 1, then y=31=3y = 3^1 = \mathbf{3}.
If x=2x = 2, then y=32=9y = 3^2 = \mathbf{9}.
If x=−1x = -1, then y=3−1=13≈0.33y = 3^{-1} = \frac{1}{3} \approx \mathbf{0.33}.
If x=−2x = -2, then y=3−2=19≈0.11y = 3^{-2} = \frac{1}{9} \approx \mathbf{0.11}.

STEP 4

From our calculations, we found the point (0,1)(0, 1).
This is the **y-intercept**!
Since 3x3^x will always be positive, the graph will *never* cross the x-axis, so there's no x-intercept.

STEP 5

As xx becomes more and more negative, 3x3^x gets closer and closer to zero.
This means the **horizontal asymptote** is y=0y = \mathbf{0}.

STEP 6

Plot the points we found and draw a smooth curve that approaches the asymptote as xx goes to negative infinity.
The graph should increase rapidly as xx increases.

STEP 7

Let's plug in some values for xx.
If x=0x = 0, then y=−3(12)0=−3⋅1=−3y = -3(\frac{1}{2})^0 = -3 \cdot 1 = \mathbf{-3}.
If x=1x = 1, then y=−3(12)1=−3⋅12=−1.5y = -3(\frac{1}{2})^1 = -3 \cdot \frac{1}{2} = \mathbf{-1.5}.
If x=2x = 2, then y=−3(12)2=−3⋅14=−0.75y = -3(\frac{1}{2})^2 = -3 \cdot \frac{1}{4} = \mathbf{-0.75}.
If x=−1x = -1, then y=−3(12)−1=−3⋅2=−6y = -3(\frac{1}{2})^{-1} = -3 \cdot 2 = \mathbf{-6}.
If x=−2x = -2, then y=−3(12)−2=−3⋅4=−12y = -3(\frac{1}{2})^{-2} = -3 \cdot 4 = \mathbf{-12}.

STEP 8

From our calculations, the **y-intercept** is (0,−3)(0, -3).

STEP 9

As xx becomes more and more positive, (12)x(\frac{1}{2})^x gets closer and closer to zero, so −3(12)x-3(\frac{1}{2})^x gets closer and closer to zero.
This means the **horizontal asymptote** is y=0y = \mathbf{0}.

STEP 10

Plot the points and draw a smooth curve that approaches the asymptote as xx goes to positive infinity.
The graph should decrease rapidly as xx decreases.

STEP 11

For y=3xy = 3^x, the y-intercept is (0,1)(0, 1), there is no x-intercept, and the horizontal asymptote is y=0y = 0.
For y=−3(12)xy = -3(\frac{1}{2})^x, the y-intercept is (0,−3)(0, -3), there is no x-intercept, and the horizontal asymptote is y=0y = 0.
The graphs are sketched as described above.

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