Need help fast!

Variable y varies directly with x^2, and y = 96 when x = 4.

What is the value of y when x = 2?

y = ?

Question

Need help fast!

Variable y varies directly with x^2, and y = 96 when x = 4.
 
What is the value of y when x = 2?
 
y =  ?

anonymous · Answer

@Hero  @Luigi0210

luigi0210 · Answer

For this specific problem your equation would be $y=kx^2 $ 

Plug in the given values to solve for k

anonymous · Answer

Um how do i find k? isnt k=y/x?
So... 384? that couldnt work...

anonymous · Answer

A little help? @Luigi0210

anonymous · Answer

@Hero @bohotness @hhelpplzzzz  @k_lynn

mathstudent55 · Answer

No.
$y = kx^2$ is the general form of quadratic variation.
A specific case can be $y = 2x^2$ or $y = \dfrac{2}{3}x^2$.
What you need to do is to find what is the value of k in your specific problem.

mathstudent55 · Answer

Start with the general form:

$y = kx^2$

We need to find the value of k, but there are 3 unknowns in the equation. The problem tells you that when x = 4, y = 96. You use the general equation and plug in 4 for x and 96 for y. Now the only unknown is k. You can now solve for k.

Once you find y, you can rewrite the equation with the correct value of k plugged in for k.

anonymous · Answer

Yeah but... how do i find k? and in this case dosent it mean the awnser is 96? @mathstudent55

anonymous · Answer

96 diveded by 2*

mathstudent55 · Answer

No.
Here it is step by step.
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mathstudent55 · Answer

Since the problem tells us y = 96 when x = 4, we replace x with 4 and y with 96.
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