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Simple X and Y intercept question - Thanks

  • Thread starter nukeman
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Homework Statement



Im doing functions, and graphing functions. Here is the question:

Find the X intercepts, and Y intercepts of

f(x) 2x^2 - x -1

Homework Equations





The Attempt at a Solution



So first finding the X intercepts, the first thing I do, is change it to:

f(x) = 0 so,

2x^2 - x -1 = 0

Now the book takes the above, and changes it to:

(2x+1)(x-1) = 0 (How do you get this <---, from 2x^2 - x -1 = 0 )

Then it goes,

x = -1/2, x = 1

So the X intercepts are (-1/2,0) and (1,0)

Any help would be great!

P.S: Can you work this out on a graphing calculator? Just curious...
 
Last edited:

Answers and Replies

  • #2
LCKurtz
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Homework Statement



Im doing functions, and graphing functions. Here is the question:

Find the X intercepts, and Y intercepts of

f(x) 2x^2 - x -1

Homework Equations





The Attempt at a Solution



So first finding the X intercepts, the first thing I do, is change it to:

f(x) = 0 so,

2x^2 - x -1 = 0
So you have a quadratic equation to solve. The two methods for solving such an equation are to factor it if possible, or use the quadratic formula.

Now the book takes the above, and changes it to:

(2x+1)(x-1) = 0 (How do you get this <---, from 2x^2 - x -1 = 0 )
Well, you can check that is correct by multiplying it out. Have you not studied factoring?

Then it goes,

x = -1/2, x = 1

So the X intercepts are (-1/2,0) and (1,0)
That uses the property of numbers that if ab = 0 then either a or b can be zero, and conversely. So if 2x +1 = 0 what is x? And if x-1 = 0 what is x?
 
  • #3
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They simply factored 2x^2 - x -1

If you notice, (2x+1)(x-1) is equal to 2x^2-x-1 after all. Just distribute (Foil) out the two factors and you get something like:
2x^2 -2x +x -1

(The next step is to use the zero product property, that's why it's important to set the polynomial to zero)

Here is a short tutorial on how to do this:
http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut7_factor.htm

When the leading coefficient of the polynomial is not 1, sometimes people use another method called the "box" method to factor. As the polynomials get more and more complicated, or simply unfactorable, you can just use the Quadratic Formula.
 
  • #4
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P.S: Can you work this out on a graphing calculator? Just curious...
Yes, but you should not need to do so. It's better to get in the practice of doing in on paper.
 
  • #5
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Yes, I got the factoring part.

I totally forogt about the zero product property! YES, got it thanks!

So, that takes care of the X-int's, how about the Y int's ?

Do I simply solve the equation?

So, for f(x) = 2x^2 - x -1

I would just go 2(0)^2 - (0) - 1

Which would be: 0

So, Y intercept would be 0,0 ?

P.s. - Mark44, I agree
 
  • #6
LCKurtz
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Yes, I got the factoring part.

I totally forogt about the zero product property! YES, got it thanks!

So, that takes care of the X-int's, how about the Y int's ?

Do I simply solve the equation?

So, for f(x) = 2x^2 - x -1

I would just go 2(0)^2 - (0) - 1
Yes, y value when x = 0 is the y intercept.

Which would be: 0
0 - 0 - 1 = 0????
 
  • #7
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lol, meant to say, 0, -1

For the Y, what is the X is not 0 and lets say 1 ? Would I just still solve out and replace the x with 1 and solve?
 
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  • #8
LCKurtz
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lol, meant to say, 0, -1

For the Y, what is the X is not 0 and lets say 1 ? Would I just still solve out and replace the x with 1 and solve?
The y intercept is the point where the graph touches the y-axis, so you need x = 0.
 

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