need help with math?

Question

i have to find the vertx, x and y intercepts and axis and then draw the graph of y=-x^2+4x

Answer 1

The easy approach: factor the quadratic:

y = x(x + 4)

Find the x-intercepts by setting it to 0:

x(x + 4) = 0

x = 0, x = -4

Making, of course, the x-intercepts (0, 0) and (-4, 0)

Halfway between these, you'll find the vertex:

(0 + (-4))/2 = -2

The y-coordinate would be

y = (-2)^2 + 4(-2) = 4 - 8 = -4

The vertex, then, is (-2, -4)

Find the y-intercept by setting x to 0:

y = 0^2 + 4(0) = 0

So the y-intercept is (0,0)

which we already knew, as it was one of the x-intercepts.

Note that this way of finding the vertex - averaging the x-coordinates of the x-intercepts - will always work, but some teachers are picky about letting you use it.

Answer 2

Hope this helps!

For parabolas, taking the derivative of this function gives it's slope function. Setting this slope = 0 gives the minimum or maximum for a parabola, depending if it opens up (positive slope) or down (negative slope).
This is positive, so the function has a minimum.

The derivative of this function is:
y' = 2x + 4,
when y' = 0, x' = -2
Plug x=-2 back into the function and solve for y.
The function has a minimum at (-2,-4)

X intercepts happen when Y=0, and Y intercepts happen when X=0 in the original function.
When X=0, Y=0 and in this case vice versa,
When Y=0, X=0

To graph the function plot a couple of points, including the vertex.
(-2,4) (0,0) (1,5) (-1,-3)

Answer 3

The eqn correponds to a parabola

now -x^2 +4x = - (x^2 - 4x)
take (x+a)^2 = x^2 + a^2 + 2*a*x
add a "-" sign then

-(x+a)^2 = - x^2 - a^2 - 2*a*x

On comparison we get
-2*a*x = -4x implies a=2

so

-(x+2)^2 = -x^2 - 4 - 4x
if we add 4 on both sides

-(x+2)^2 + 4 = -x^2 -4x this is = y

so y = 4 - (x+2)^2
or y - 4 = -(x+2)^2

so vertex = 2 , 4

4 cuz y-4=0
and 2 cuz x+2=0 gives -2 for the "-" sign before(in eqn) it is 2

----- X intercept-------
put y=0

-x^2+4x=0 gives x(x-4)=0 gives x=0 or 4

-----Y intercept-------
put x=0
gives y=0

-------GRAPH IT!!-------------

join the points

(2 ,4) (0,0) (4,0)

done!!!!!!!

here is the link to graph
http://wims.unice.fr/wims/wims.cgi?session=popup&wims_window=550x400&module=tool%2Fgeometry%2Fanimtrace.en&precision=12&cmd=new&type=explicit2D&coord=cartesian&set_border=1&set_zeroaxis=1&set_grid=0&set_tics=1&set_transparent=0&xsize=300&ysize=270&y1=4*x-x*x&xleft=-10&xright=10&yleft=-10&yright=10

click on it

Answer 4

You find the y intercept by plugging in 0 for x.

y = 0^2 + 4(0) = 0.

The y intercept is 0.

Similarly you find the x intercept by plugging in 0 for y and solving.

0 = x^2 + 4x
0 = x(x + 4)
0 = x, 0 = x + 4
x = -4, x = 0

The x intercept is -4 and 0.

Vertex is the derivative set to 0.

0 =2x + 4
2x = -4
x = -2

The vertex is -2.

Answer 5

y=-x^2+4x
y=-x^2+4x+4-4= (x^2+4x+4) -4
=(x+2)^2 -4
The vertex is (-2,-4)
To find x intercept put y=0 therefore x^2+4x=0 or x(x+4)=0, therefore x intercepts are 0 and -4
To find y intercept put x=0,then y=0
It means the curve or parabola passes through (0,0) and (0,-4)
the ax-ix of the curve is x=-2
The curve you have to draw yourself

Answer 6

vertex is 2
x intercept is when y equals 0 now is 0
y intercepts are when x equals 0 wich is 0
so it just comes from the origin and goes up exponetialy

Answer 7

vertex(r,k)

r=-b/2a=-4/-2=2

k=(4ac-b^2)/4a=4*(-1)*0-16/-4=4

for x intercept y=0
0=-x^2+4x
x(-x+4)=0
Hence
x1=0
x2=4

For y intercept
x=0
y=-0^2+4*0=0

The arms are down because of minus sign.

Answer 8

for any quadratic, use the equation -b/2a to find the x value of the vertex, if I remember correctly. :) The rest you can figure out from there!

Answer 9

be careful, this is a nonconservative field in the complex plane so you can NOT use Gauss's Law

Answer 10

Well the answer is very simple.

Use the formula Y=MX+B.

trust me. it may not make sense but you will definetly get the answer.