Question

22.
x | 0 | 1 | 3 | 4
y | 5 | 15 | 25 | 35
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: exponential decay, linear growth, quadratic (local max), linear decay, exponential growth

Explanation:

Step1: Calculate the differences between consecutive y - values

For \(x = 0,y = 5\); \(x = 1,y = 15\); \(x = 3,y = 25\); \(x = 4,y = 35\)
The differences are \(15 - 5=10\), \(25 - 15 = 10\), \(35 - 25=10\)

Step2: Check the nature of the function

Since the difference between consecutive \(y\) - values is constant (\(\Delta y=10\)), and the \(x\) - values have a constant difference (\(\Delta x = 1\) for \(x = 0\) to \(x = 1\), \(\Delta x=2\) for \(x = 1\) to \(x = 3\) but when we calculate the rate of change \(\frac{\Delta y}{\Delta x}\), for \(x = 0\) to \(x = 1\), \(\frac{15 - 5}{1-0}=10\); for \(x = 1\) to \(x = 3\), \(\frac{25 - 15}{3 - 1}=\frac{10}{2} = 5\) (incorrect approach, better to use the formula for a linear function \(y=mx + b\)). A linear function has the form \(y=mx + b\), where \(m\) is the slope. Using two points \((x_1,y_1)=(0,5)\) and \((x_2,y_1)=(1,15)\), \(m=\frac{y_2 - y_1}{x_2 - x_1}=\frac{15 - 5}{1-0}=10\) and \(b = 5\) (when \(x = 0,y=b\)). The function is \(y = 10x+5\). For a quadratic function \(y=ax^{2}+bx + c\), the second - differences are constant. For an exponential function \(y = a\cdot b^{x}\), the ratios of consecutive \(y\) - values are constant.

Answer:

Linear Growth