how do you graph this?

Question

3) f(x)= -1/3x^2
4) f(x)=x^2-6x+9

Answer 1

Carefully, very carefully :)

f(x) =-1/3x^2 is a parabola origin at (0,0) with branches downward.


f(x)=x^2-6x+9 =(x-3)^2 orgin at x=3, y intecept at x=0

Answer 2

I'm assuming you have taken Calculus.

3) f(x)= -1/3x^2
f(x)= (-1/3)*x²
f'(x)=(-2/3)*x
f''(x)=(-2/3)
Solve for f'(x)=o
=> x=0
Is at x=0 a min or max?
Since f''(0)= -2/3 =>
at x=0 is a max.
Note: f''(xo)<0 => max
if f''(xo)>0 => min pt at xo.

Make a chart of x & y values.
at x=0 => y=0
at x=+- 1 => y= -1/3
at x=+-2 => y= -4/3 etc.
then plot on graph paper
x & y values. You have a parabola
that 'spills water' where the max point is
at x=0 & y=0. It looks like an upside U with the max at x=0 and all other points in the third & fourth quads on x-y axis. No points above the line y=0, all of your graph is below the st line y=0. I can't draw on the keyboard.

4) f(x)=x^2-6x+9
Again as before, you will look for min or max points for the parabola.
f'(x)=2*x-6
f''(x)=2
----------------------
Note: f''(xo)<0 => max
if f''(xo)>0 => min pt at xo.
------------------------
solve for x where
f'(x)=2*x-6=0
2x=6
x=3
f''(3)= 2
since f''(3) is positive
then you have a min point at
x=3. Parabola & Holds water U.
-----------------------------
solve x²-6x+9=0 for x
x={6+-sqrt(36-36)}/2
x=3 is only Real Root of eq
Plus this is a min point.
Make a chart
y=x^2-6x+9
x=3 & y=9-18+9=0
At x=3 & y= 0
you have a minimum pt.
make chart of x & y's
x= 2 & y= 4-12+9=1
x= 4 & y= 16-24+9=1
x= 1 & y=2-6+9=5
x=5 & y=25-30+9=4

Plot the points on graph paper
You have parabola tha holds water
that looks lie U
with min point at x=3 & y=0.
----------------------------

Answer 3

Look at the generic form y = aX^2 + bX + c.

The vertex will always be at -b/(2a). In the first function, the b and c terms are zero, so it must be a parabola with its vertex at x = 0. Because the 'a' term is negative, it will open downward.

For the second one, -b/(2a) = -(-6)/(2*1) = 3. It will have it's vertex at x = 3. Plug 3 into the function and you get back y = 0. We would have seen this quickly if we had looked at the factored form of f(x) = x^2-6x+9 = (x-3)^2.

This would have told us that it looks just like y = x^2, but moved to the right 3. So, yes, the second parabola opens up, has it's vertex at (3,0), and it passes through (0,9).

Answer 4

3.)
f(x) = (-1/3)x^2
unless you mean
f(x) = -1/(3x^2)

4.)
f(x) = x^2 - 6x + 9

For a graph, go to www.calculator.com/calcs/GCalc.html

type in (-1/3)x^2 click enter, then erase it and type in x^2 - 6x + 9. Make sure to leave out the f(x) part. Also notice that i put (-1/3) into a parenthesis, its not only because its a fraction, but also because its negative.

Answer 5

Use Cartesian diagram (x,y diagram). Make the f(x) value as its y value (vertical axis), and x into x value (horizontal axis). Make some points from those intercept values by put some values into x and than latter connect those points.

For example:
f(x) = x^2-6x+9
put x = -1, f(x) = y = 16
x = 0, y = 9
x =1 , y = 4
so on

Note: make points as much as possible to help you make the graph easier.

Answer 6

They are both parabolas because they are quadratic functions, the first one is upside down and thinner, the second one is upright

Answer 7

you wnat to plot them..just plot then..they are parabolic curves..

Answer 8

http://www.coolmath.com/graphit/index.html

this should be help ful