H(x) = 3x2 - kx nd h(4) = h(-2) , then find vaue of k

[Riwaj]

Given that h(x) = 3x² - kx nd h(4) = h(-2) , then find vaue of k .

 

[MarkFL]

Given that h(x) = 3x² - kx nd h(4) = h(-2) , then find vaue of k .

We are given that:

\(\displaystyle h(x)=3x^2-kx\)

And so:

\(\displaystyle h(4)=3(4)^2-k(4)=?\)

\(\displaystyle h(-2)=3(-2)^2-k(-2)=?\)

Simplify the above, and then equate the two expressions, because we are told $h(4)=h(-2)$, and you will be able to solve for $k$. :)

Another way to proceed would be to observe that the axis of symmetry of this quadratic polynomial must be:

\(\displaystyle x=\frac{4+(-2)}{2}=1\)

Given that for the general quadratic $ax^2+bx+c$, the axis of symmetry is at:

\(\displaystyle x=-\frac{b}{2a}\)

Equate the two values for the axis of symmetry, and solve for $k$. :)

 

[Riwaj]

hi mark fl

 

[MarkFL]

hi mark fl

Hello! (Wave)

 

[Riwaj]

sir , may i get you email please ?

 

[MarkFL]

sir , may i get you email please ?

I generally don't give out my email address unless it is necessary for some reason. You can contact me via our private message system here. :)

You may have noticed that I moved your thread from the Trigonometry forum to our Algebra forum, as that's a better fit for the problem, and I gave the thread a title that describes the problem. Putting threads in the best forum suited to the problems being discussed, and giving threads titles that briefly describe the problem are things we ask our users to do for better forum organization, as indicated in our forum rules. :)

 

[Riwaj]

Oh sorry sir , the thing is that i am new in this forum . So, i don't know much about it and i frequently get confused . I will try my best to not repeat it any time .

 

[MarkFL]

Oh sorry sir , the thing is that i am new in this forum . So, i don't know much about it and i frequently get confused . I will try my best to not repeat it any time .

That's quite understandable, and I don't mean to give the impression that I am scolding or chiding you in any way...I only mention these things for future reference. There's always a learning curve with any site, and we are no exception to this. :)

 

[Riwaj]

got it sir , and sir ease solve my problem in the algebra forum .
the thing is that today is the last day of my vacation and tomorrow i have to submit my opt. maths homework . so its very urgent .

 

[MarkFL]

...the thing is that today is the last day of my vacation and tomorrow i have to submit my opt. maths homework . so its very urgent .

So, you waited until the last day of your vacation to begin your homework? (Sweating)(Lipssealed)

As I have told students in the past..."Your procrastination does not become our emergency." (Nerd)

We are happy to help, but we do actually help rather than simply complete problems for people. And we don't rush because someone put off their homework until the last minute. Haste makes waste. (Speechless)

 

[MarkFL]

You've indicated that the problem has been solved, but you didn't follow up by posting your work. Doing so shows the helper that you've understood the help provided, and it makes the thread more useful to those who've found this thread via a search and who are wanting to see how such a problem can be worked.

We are given that:

\(\displaystyle h(x)=3x^2-kx\)

And so:

\(\displaystyle h(4)=3(4)^2-k(4)=?\)

\(\displaystyle h(-2)=3(-2)^2-k(-2)=?\)

Simplify the above, and then equate the two expressions, because we are told $h(4)=h(-2)$, and you will be able to solve for $k$. :)

\(\displaystyle 48-4k=12+2k\)

\(\displaystyle 6k=36\)

\(\displaystyle k=6\)

Another way to proceed would be to observe that the axis of symmetry of this quadratic polynomial must be:

\(\displaystyle x=\frac{4+(-2)}{2}=1\)

Given that for the general quadratic $ax^2+bx+c$, the axis of symmetry is at:

\(\displaystyle x=-\frac{b}{2a}\)

Equate the two values for the axis of symmetry, and solve for $k$. :)

\(\displaystyle -\frac{-k}{2(3)}=1\)

\(\displaystyle k=6\)