help plz again (2 Viewers)

[super.muppy]

what is the maximum value of the function y=15-(x-4)^2 and for wat value does it occur?
how do u do dat
i dnt even understand it

 

[Aznmichael92]

to see if u get a maximum value or a minimum value, u have to know whether its a concave up or concave down parabola. If you look at the sign in front of the x^2 thing, it will tell you whether its a negative or positive graph. Positive graph are concave up thus resulting in a minimum point. Negative graph are concave down thus resulting in a maximum point.

This graph is negative and you will be looking for a maximum point as there is a - sign in front of (x-4)^2

We expand brackets so we can use vertex formula. The one like -b/2a.

y = 15-(x-4)^2
= 15 - (x^2 - 8x + 16)
= 15 - x^2 + 8x - 16
= -x^2 + 8x - 1

so now we fine the vertex (point x on minimum or maximum point)

x=-(8)/2(-1)
= -8/-2
= 4

now we sub x into equation to get point.

y=15 - (4-4)^2
= 15 - 0^2
= 15

therefore maximum point is (4,15)

I hope that helped and hope its correct. I am scared...

 

[rolror88]

I'm pretty sure michael did it right.

 

[Aznmichael92]

I'm pretty sure michael did it right.

hooorah, someone thinks i am correct.

 

[super.muppy]

to see if u get a maximum value or a minimum value, u have to know whether its a concave up or concave down parabola. If you look at the sign in front of the x^2 thing, it will tell you whether its a negative or positive graph. Positive graph are concave up thus resulting in a minimum point. Negative graph are concave down thus resulting in a maximum point.

This graph is negative and you will be looking for a maximum point as there is a - sign in front of (x-4)^2

We expand brackets so we can use vertex formula. The one like -b/2a.

y = 15-(x-4)^2
= 15 - (x^2 - 8x + 16)
= 15 - x^2 + 8x - 16
= -x^2 + 8x - 1

so now we fine the vertex (point x on minimum or maximum point)

x=-(8)/2(-1)
= -8/-2
= 4

now we sub x into equation to get point.

y=15 - (4-4)^2
= 15 - 0^2
= 15

therefore maximum point is (4,15)

I hope that helped and hope its correct. I am scared...

thanx im stupid

 

[super.muppy]

hey but u guys got any notes on any topics for test

 

[Aznmichael92]

hey but u guys got any notes on any topics for test

wat do u mean by notes? formulas? btw u r not stupid. Dont think that. Its probably the thing wasnt explained properly to you. My god my sister was teachinng me the other day, she was teaching for 10 minutes i still didnt understand. I was like I am lost. 10 minutes later she said something that I said and I was right. OMG

 

[super.muppy]

lik notes for test lik science, history, geo, etc. i also really need engineering notes
teacher onli does prac so we learn nothing

 

[tommykins]

回复: Re: help plz again

michael is correct.

 

[Aznmichael92]

Re: 回复: Re: help plz again

michael is correct.

MUHAHAHAHAHA have another person to support me

http://community.boredofstudies.org/showthread.php?t=178765

lik notes for test lik science, history, geo, etc. i also really need engineering notes
teacher onli does prac so we learn nothing

look through this thread, it should help

 

[kaz1]

to see if u get a maximum value or a minimum value, u have to know whether its a concave up or concave down parabola. If you look at the sign in front of the x^2 thing, it will tell you whether its a negative or positive graph. Positive graph are concave up thus resulting in a minimum point. Negative graph are concave down thus resulting in a maximum point.

This graph is negative and you will be looking for a maximum point as there is a - sign in front of (x-4)^2

We expand brackets so we can use vertex formula. The one like -b/2a.

y = 15-(x-4)^2
= 15 - (x^2 - 8x + 16)
= 15 - x^2 + 8x - 16
= -x^2 + 8x - 1

so now we fine the vertex (point x on minimum or maximum point)

x=-(8)/2(-1)
= -8/-2
= 4

now we sub x into equation to get point.

y=15 - (4-4)^2
= 15 - 0^2
= 15

therefore maximum point is (4,15)

I hope that helped and hope its correct. I am scared...

That's a bit tedious. You can just say that since the parabola is in the form y=k+or-(x-h)² where (h,k) is the vertex. I wrote plus and minus because it doesn't matter.

 

[Aznmichael92]

That's a bit tedious. You can just say that since the parabola is in the form y=k+or-(x-h)² where (h,k) is the vertex. I wrote plus and minus because it doesn't matter.

is that the quicker way of doing it? OMG I AM SO NOOOB!!!

 

[Aznmichael92]

um u do realise he is in year 10 and i am also in year 10? Even if i knew how to do calculus might not be a good idea to explain to someone who hasnt learnt the topic?

 

[minijumbuk]

You can actually just look at the equation.

y = 15 - (x-4)^2
Since you know for all values of x, the square will always make it positive, then you know that for any value of x (besides x=4), y < 15.
So, the max value is when (x-4)^2 = 0
So max value is y = 15 when x = 4