Chapter 9.9, Problem 31HP

To determine

A function

Answer to Problem 31HP

The function is $y=2x^2-5$

Explanation of Solution

Given:

Second differences, first differences are no constant, a y-intercept of $-5$ and passes through the point $\left(2,3\right)$

Concept Used:

If first difference of y-values is equal, then model be linear

If second difference be equal, then the model be quadratic

If first and second differences are not equal then the model will be exponential

Calculation:

In order to find a function

If the second difference are not constant but the first difference are not constant, then the function is a quadratic and y- intercept is not 0, the function will have the form

  $y=a{x}^2+c$

Use the point $\left(2,3\right)$ to find the value of a for the function

  $\begin{array}{c}\left(2,3\right)\\ c=-5\\ y=a{x}^2+c\\ 3=a{\left(2\right)}^2-5\\ 3=4a-5\\ 3+5=4a-5+5\\ 8=4a\\ \frac{8}{4}=\frac{4a}{4}\\ 2=a\\ \Rightarrow a=2\end{array}$

Thus, the function is $y=2x^2-5$