Help Identifying x-intercepts of quadratic functions

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SUMMARY

The discussion focuses on identifying the x-intercepts of the quadratic function f(x) = 1/2x² - 4. The user initially attempted to factor the equation but struggled with the process. A participant clarified that factoring is unnecessary for finding the x-intercepts and provided a straightforward solution: setting (1/2)x² - 4 = 0 leads directly to x² = 8, resulting in x = ±2√2 as the x-intercepts.

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Kalzar89
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Homework Statement


Identify the vertex and x-intercept(s) of the quadratic function algebraically.



Homework Equations



f(x) = 1/2x^2 - 4

1/2 Is the coefficient if its not clear.



The Attempt at a Solution



Ok I do not need help finding the vertex I have already got that, I need help factoring it out to find the x-intercepts.

Here is my attempt

0 = 1/2x^2 - 4
0 = 1/2 (x^2 - 8)

That is as far as I can factor it, I am terrible at factoring these and I actually have a way easier time factoring trinomials and polynomials.

Could you please explain how I am to factor farther and how am I supposed to solve for the x-intercepts after its all factored. I have even looked all over the internet for a good factoring tutorial guide but I just cannot even understand them.

Thanks
 
Last edited:
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Kalzar89 said:
Ok I do not need help finding the vertex I have already got that, I need help factoring it out to find the x-intercepts.

Here is my attempt

0 = 1/2x^2 - 4
0 = 1/2 (x^2 - 8)

That is as far as I can factor it, I am terrible at factoring these and I actually have a way easier time factoring trinomials and polynomials.

There is no need to worry that much about factoring. You have

\frac{1}{2}x^{2}-4=0.​

Simply solve for x to obtain the intercept.
 
I believe that was what he was trying to do! Kalzar89, there is no need to factor.
(1/2)x^2- 4= 0
(1/2)x^2= 4
x^2= 8
and take the square root of both sides.
x= \pm\sqrt{8}= \pm \sqrt{(4)(2)}= \pm 2\sqrt{2}
 

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