Can't seem to get this graphing question. (1 Viewer)

[girlworld_club]

Not sure if the back of the book is wrong.

Sketch y=6-2x-x^2 showing the stationary point. What i keep getting wrong is the x intercepts on the graph. I used quadratic formule, tried factorising but nothing works.

 

[iBibah]

http://www.wolframalpha.com/input/?i=6-2x-x^2+=0

 

[girlworld_club]

i was right. stupid good for nothing book.

 

[girlworld_club]

Oh and thanks

 

[girlworld_club]

Also how do you find the stationairy points for:
y = (x-4)(x+2)^2

i don't get the right answer, and i know i am doing something wrong.

 

[Nws m8]

Also how do you find the stationairy points for:
y = (x-4)(x+2)^2

i don't get the right answer, and i know i am doing something wrong.

Differentiate and whatever you get, make it equal to zero ....and then find x and y values and that is stationary point(s)

And did you get the chem question on your other thread?

 

[iBibah]

Also how do you find the stationairy points for:
y = (x-4)(x+2)^2

i don't get the right answer, and i know i am doing something wrong.

Product Rule:

u=(x-4)
v=(x+2)^2

 

[girlworld_club]

Guys update: for the graph, i got a stationairy point as a point of inflextion is this possible?
y = (x-4)(x+2)^2

the graph looks retarded can someone graph it showing working as to how they got their answer. Please include points of inflection.

 

[iBibah]

Guys update: for the graph, i got a stationairy point as a point of inflextion is this possible?
y = (x-4)(x+2)^2

the graph looks retarded can someone graph it showing working as to how they got their answer. Please include points of inflection.

Yes it is. Horizontal point of inflexion. However this graph does not have one.

http://www.wolframalpha.com/input/?i=y+=+(x-4)(x+2)^2