Chapter 2.2, Problem 2E

To determine

To Calculate: The quadratic equation that belongs to other three.

Answer to Problem 2E

  $f(x)=3x^2+24x-6$

Explanation of Solution

Given information: In the question four quadratic functions are given, we have to find which one does not belong with other three.

Formula used: we know that any quadratic function in vertex form is $y=a{(x-h)}^2+k$ .

Calculation: If I take one by one all quadratic function and write in vertex form and then compare.

                              1.                                    2.                                                            3.                                          4.      

  $\begin{array}{}$ f(x)=3x^2+6x-24$$ f(x)=3\left(x^2+2x-8\right)$$ f(x)=3\left\{{(x+1)}^2-9\right\}$$ f(x)=3{(x+1)}^2-27$\end{array}$  $\begin{array}{}$ f(x)=3x^2+24x-6$$ f(x)=3\left(x^2+8x-2\right)$$ f(x)=3\left\{{(x+4)}^2-18\right\}$$ f(x)=3{(x+4)}^2-54$\end{array}$  $\begin{array}{}$ f(x)=3(x-2)(x+4)$$ f(x)=3x^2+6x-24$$ f(x)=3\left(x^2+2x-8\right)$$ f(x)=3\left\{{(x+1)}^2-9\right\}$$ f(x)=3{(x+1)}^2-27$\end{array}$  $f(x)=3{(x+1)}^2-27$

From this comparison we get only option 2.is different from other three. Hence we can say that $f(x)=3x^2+24x-6$ is different from other means does not belong to other three.