Question

Given: f(x) = 2(x + 1)2 - 51)what direction does the graph open2)maximum or minimum3)name the domain and range4) write the equation in standard formI already graph itvertex is 2,5

Answer 1

We have the function:

`f(x)=2(x+1)^2-5`

We can calculate the minimum value of f(x) looking at (x+1)^2.

This part of the function has a minimum value of 0 (when x=-1). For any other value of x, this part of the function is always higher than 0.

Then, if this is the minimum value for (x+1)^2, then the minimum value for f(x) is f(-1)=0-5=-5.

We can express f(x) as:

`\begin{gathered} f(x)=2(x+1)^2-5 \\ f(x)=2(x^2+2x+1)-5 \\ f(x)=2x^2+4x+2-5 \\ f(x)=2x^2+4x-3 \end{gathered}`