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consider the function $f(x)=8 - 6(0.75)^x$. its graph is shown below al…

Question

consider the function $f(x)=8 - 6(0.75)^x$. its graph is shown below along with its horizontal asymptote. why does this function increase to its horizontal asymptote instead of decrease like others in this lesson?
(graph: $y = 8$ (horizontal line), $y = 8 - 6(0.75)^x$ (curve), axes with y from 0 to 10, x from 0 to 15 approximately)
n-gen math® algebra ii – unit 4 – exponential functions – lesson 9

Explanation:

Step1: Analyze the exponential term

The function is \( f(x) = 8 - 6(0.75)^x \). The base of the exponential term, \( 0.75 \), is between \( 0 \) and \( 1 \), so \( (0.75)^x \) is a decreasing exponential function (as \( x \) increases, \( (0.75)^x \) decreases towards \( 0 \)).

Step2: Analyze the effect of the negative coefficient

We have a negative coefficient (\( -6 \)) in front of the exponential term. So, \( -6(0.75)^x \) will be the opposite of \( 6(0.75)^x \). Since \( (0.75)^x \) decreases as \( x \) increases, \( -6(0.75)^x \) will increase (because we're multiplying a decreasing quantity by a negative number, which reverses the direction of change).

Step3: Analyze the overall function

The function \( f(x) = 8 + (-6(0.75)^x) \). As \( x \) increases, \( -6(0.75)^x \) increases (from a more negative value towards \( 0 \)), so adding this to \( 8 \) means \( f(x) \) increases towards the horizontal asymptote \( y = 8 \) (since as \( x \to \infty \), \( (0.75)^x \to 0 \), so \( f(x) \to 8 - 0 = 8 \)). In typical exponential decay functions without the negative sign (like \( f(x)=a - b(c)^x \) with \( c < 1 \) but no negative on the exponential term), the exponential term decreases, making the whole function decrease, but here the negative sign flips the behavior of the exponential part, causing the function to increase.

Answer:

The function \( f(x)=8 - 6(0.75)^x \) increases to its horizontal asymptote because the exponential term \( (0.75)^x \) (with base \( 0 < 0.75 < 1 \), so it decreases as \( x \) increases) is multiplied by \( -6 \). This makes \( -6(0.75)^x \) increase as \( x \) increases (since a decreasing positive quantity multiplied by a negative becomes an increasing quantity). Then, adding \( 8 \) to this increasing term means \( f(x) \) increases towards \( y = 8 \) (the horizontal asymptote) as \( x \) increases.