Polynomial and Rational Functions

[rgmiller8]

I am not understnding how to start this problem, could you please help me?? thanks so much!!

f(x)=2(x+2)-1

my directions read usethe vertex and intercepts to sketch the graph of each quadratic function. Give tequation of the parabola's axis of symmetry. Use the graph to determine the function's domain and range.

 

[galactus]

This is not a quadratic. It is linear.

\(\displaystyle 2(x+2)-1=2x+3\)

But, assuming you inadvertently omitted the ^2 and you really meant:

\(\displaystyle 2(x+2)^{2}-1\)

If this is the case, then it is a quadratic and represents a parabola.

It is a parabola shifted 2 units left of the origin. This we can tell from the x+2 in the parentheses.

If it were x-2, then it would be shifted 2 units right of the origin.

The -1 means it is shifted 1 unit down from the origin

The 2 is how 'spread out' it is.

The axis of symmetry is the vertical line running up the center of the parabola.

The vertex is easy to spot since the parabola is 2 units left and 1 unit down from the origin.

To find the x intercepts, set the equation equal to 0 and solve for x.

The y intercept can easily be found by setting x=0 in the given equation.

Expand it out, if you wish, into the form \(\displaystyle ax^{2}+bx+c\).

The vertex is \(\displaystyle \left(\frac{-b}{2a}, \;\ c-\frac{b^{2}}{4a}\right)\).

The coordinates of the vertex should match the results from above (the shifting).