What are the points if we graph this quadratic equation f(x) = -x^2 + 4x?

Answer 1

An instructor always wants to know the following about a graph:

The x-intercepts, the y-intercepts, and for parabolas, where is the vertex.

Set x=0 to find the y intercept. Hmm (0,0)

Set y=0 and solve to get the x-intercepts:
It factors into x(4-x) so, y=0 when x=0 and 4.

You can always get the vertex by using x= -b/(2a) or, if there are x-intercepts, it will always be halfway in between them. In this case, the vertex is at x=2. Plug that in to get y=4.

So, the important info is: It opens downward. The vertext is at (2,4). The x-intercepts are at (0,0) and (4,0) and the y-intercept is at (0,0).

Answer 2

Are you kidding?

1. f(x)=0 then the roots are x1=0, x2=4
2. The branches are downward(-1 before x^2)
3. Symmetry of parabola : x=2 f(2)=4

I hope that does it

Answer 3

-x^2 + 4x =
-b +- \/^^^b^2-4ac / 2a =
-4 +- 16-4*-1*0 /-2 =
X1 = -4 -2 = -6
X2 = -4 +2 = -2

Answer 4

If you graph the equation you will get tan. shape, and it intersect with Y-axis in 0, and with X-axis in 4.
And now, can you tel me what's the shape of this equation:
r=1-sin(a)
With regards.

Answer 5

quadratic equation= -b+/- square root(b^2-4ac)/2a

answer=0 and -4

Answer 6

f(x) = -x^2 + 4x
f(x) = -x(x - 4)

x = 0 and 4

for a graph, go to http://www.calculator.com/calcs/GCalc.html

Answer 7

Plug it into your graphing calculator and then hit 2nd table and look at the table.

=)

Answer 8

x intercepts are -4 and 0.

Points: (0,0), (1,5), (-1,-3), (-2,-4), (-4,0)

Answer 9

{(x,y): y= -x^2 + 4x}