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QUESTION IMAGE

26. | x | y | | 0 | 27 | | 1 | 9 | | 2 | 3 | | 4 | 1 | model (circle on…

Question

26.

xy
027
19
23
41

model (circle one):
linear growth, quadratic (local max), exponential growth
linear decay, quadratic (local min), exponential decay
identify whether the function is linear, quadratic, or exponential?
options: quadratic (local max), exponential growth, linear decay, linear growth, exponential decay

Explanation:

Step1: Recall function properties

Linear functions have constant rate of change ($\frac{\Delta y}{\Delta x}$). Quadratic functions have a second - order polynomial form ($y = ax^{2}+bx + c$) and their rate of change of rate of change is constant. Exponential functions have the form $y=a\cdot b^{x}$.

Step2: Check for exponential function

For an exponential function $y = a\cdot b^{x}$, when $x = 0$, $y=a$. Here, when $x = 0$, $y = 27$, so $a = 27$.
When $x=1$, $y=9$. Substitute into $y = 27\cdot b^{x}$: $9=27\cdot b^{1}$, then $b=\frac{9}{27}=\frac{1}{3}$.
Check for $x = 2$: $y=27\cdot(\frac{1}{3})^{2}=27\cdot\frac{1}{9} = 3$.
Check for $x = 4$: $y=27\cdot(\frac{1}{3})^{4}=27\cdot\frac{1}{81}= \frac{1}{3}\approx0.33$ (but if we assume the table has a typo and consider the general form). The general form of an exponential decay function is $y = a\cdot b^{x}$ where $0\lt b\lt1$. Here $a = 27$ and $b=\frac{1}{3}$

Answer:

Exponential Decay