# Functions?

• Jul 14th 2007, 02:22 AM
Dannblood
Functions?
Hey, just working on some functions and ive got a couple ive never seen before. I need to know how to get the numbers to draw the function and the domain and range.

y = (x - 2)^2 - 5

y = 3x^2 - x - 2

Any help is appreciated.
• Jul 14th 2007, 08:03 AM
Jhevon
Quote:

Originally Posted by Dannblood
Hey, just working on some functions and ive got a couple ive never seen before. I need to know how to get the numbers to draw the function and the domain and range.

y = (x - 2)^2 - 5

y = 3x^2 - x - 2

Any help is appreciated.

Parabolas are nice to graph. you need AT MOST 4 points to graph things like these.

To graph a parabola we need to find the x-intercepts, the y-intercepts, and the vertex.

To find the x-intercepts, we replace y with zero and solve for x.

To find the y-intercepts, we replace x with zero and solve for y

To find the vertex:

If the parabola is given in the form: $f(x) = y = ax^2 + bx + c$

the x-value for the vertex is given by: $x = \frac {-b}{2a}$

Thus the co-ordinate for the vertex is: $\left( \frac {-b}{2a}, f \left( \frac {-b}{2a} \right) \right)$

If the parabola is given in the form: $f(x) = y = a(x - h)^2 + k$

Then the vertex is given by: $(h,k)$

once you have these points, plot them on your graph and draw a curve thought them that looks like a parabola. Try the problems, if you need more help someone will help you
• Jul 14th 2007, 10:41 AM
starswept
Jhevon's given you lots of great info. Like he said, parabolas are easy to graph once you have the hang of them. When you're new to them, they're especially simple when you have them in standard form: y = a(x-h)^2 + k, like your first example. When graphing this type of parabola, it's best to simply know your basic points for y = x^2. Those are of course: (0,0), (1,1), (2,4), (3,9), (4,16), (-1,1), (-2,4), (-3,9), (-4,16) and so on. Then you can plot these points (I usually do the first three) and draw the basic parabola, which is y = x^2.

From there, you know that h represents a horizontal shift: -h shifts it h units to the right, and +h shifts it h units to the left. The k represents vertical shifts: +k shifts it k units up, and -k shifts it k units downwards. So, simply move each point appropriately, then connect those new points as a parabola.

For example, with your first function, y = (x - 2)^2 - 5, you would plot your basic points. Then shift 0 two units right and down 5, to get your vertex of (2,5). Then you would move (1,1) two units right and down 5, as well as (-1,1) two units right and down 5. Keep doing this to all your basic points, erase your original graph, and connect your new points as a parabola. This sounds lengthy, but is actually very very quick once you know your basic graph, and it's a good way to help you see how vertical and horizontal translations work if you're new to parabolas. (Remember that if you'd had an a value in front of that (x-2)^2, you'd also have to apply the vertical and horizontal expansion or compression, but as we don't have that here we don't need to worry about it now)

For your second example, y = 3x^2 - x - 2, your best bet would be to complete the square so that you have your equation in a(x-h)^2 + k form, and then you can graph it as explained. Have you learned to complete the square?

EDIT: Oh yes, and as far as your domain and range, the domain of a parabola is always {x|XER} (x is an element of all real numbers). If the parabola opens upwards, the range will be y is greater than or equal to the k value of your function. If it opens downwards, the range will be y is less than or equal to the k value.