Quadratic functions (from Guide 3) - position of parabola?

[bradobrown]

Question refers to page 171 of:

Sub-topic - Quadratic Functions;
Topic - Optimization Problems section;
Chapter - Formula & Functions - Advanced Strategy (Ch.11);
Guide Book - Book 3 - Equations, Inequalities & VICs (4.1 edition).

The Manhattan book explains how to find the min or max values of quadratic functions really well but I am slightly unsure as to how the parabola for the 5th function, at the bottom of page 171, f(x) = 2 + (x+3)^2 is formed/positioned.

Why is the U-shaped parabola only draw in the top left-hand corner of the graph i.e. only negative x-values have been plotted? Why have there been no values for positive x plotted? All the other graphs above this one have both negative and positive values plotted.

Using the following negative values for x, one should be able to draw the graph relevant to the function f(x) = 2 + (x+3)^2:

if x=0, y=10
if x=-1, y=6
if x=-2, y=3
if x=-3, y=2
if x=-4, y=3
if x=-5, y=6
if x=-6, y=11

The above values seem to follow the shape and location of the parabola in the book.

But if x is positive, the shape of the graph appears entirely different:

if x=0, y=10
if x=1, y=18
if x=2, y=25
if x=3, y=38
if x=4, y=51
if x=5, y=68
if x=6, y=81

This seems more like an ever-increasing linear graph than a parabola.

I would be really grateful if you could point out what information I am over-looking as something must be erroneous somewhere.

[bradobrown]

Can anyone provide some assistance please?

[StaceyKoprince]

Redraw the graph on your scrap paper. See the ends of the two lines where the arrows are? Keep extending them out. The arrows indicate that the parabola keeps going in those directions.

So, if you draw it out further, you will get the additional points that you calculated - the graph in the book just doesn't draw it out that far. Note that the additional points you calculated are only ONE side of the parabola. The minimum occurs at x=-3 and all other values of x return higher values of y, so it increases continuously and forever, from the minimum - just like the other parabolas.

[bradobrown]

Great explanation Stacey, got it in one! Thanks a lot, it was driving me crazy.

I've just realized that I made an error in one of the y calculations up top. I wrote: "for function f(x) = 2 + (x+3)^2: if x=0, y=10". It should of course be if x=0 then y=11.

P.S. Sorry for replying to my own thread, I was worried it was going to remain unanswered. Hope you had a great break.

[tim]

Glad to hear this one makes sense now. Remember of course that the best way to guarantee your thread never gets answered is to keep replying to it.. :)