The graph of which functions has an axis of symmetry at x = -1/4?

2x^2 + x - 1

If I did anything you didn't understand let me know so I can explain.  

Step-by-step explanation:

All of them are quadratics so let's use that.

The first one is 2x^2 + x - 1.  To find the axis of symmetry the strategy is usually to find the two zeroes of a quadratic and pick the number between them.  Something to  notice though is that 2x^2 + x - 1 is just 2x^2 + x sshifted down 1, so they both have the same axis of symmetry.  So I am going to ignore the constant, because then finding the zeroes is much much simpler.  I am going to do this with all opions.

So 2x^2 + x - 1 I am just going to use 2x^2 + x.  If you factor out an x you get x(2x + 1)  So now we have it in factored form and we know the zeroes are 0 and -1/2.  The number directly in between these is -1/4, so the axis of symmetry is x = -1/4.  I don't know if there is only one with that axis of symmetry so i am going to check the rest.

2x^2 - x + 1 means we are only going to look at 2x^2 - x.  factoring we get x(2x - 1) so the zeroes are 0 and 1/2, so the axis of symmetry is at 1/4.

x^2 + 2x - 1 we only use x^2 + 2x.  Factored form is x(x+2) so zeroes are 0 and -2 whichh means axis of symmetry is -1

x^2 - 2x + 1 has the same axis of symmetry as x^2 - 2x, which has zeros at 0 and 2 so the axis of symmetry is at 1.

So yep, it was just the first one.

9514 1404 393

  (a)  f(x) = 2x^2 +x -1

Step-by-step explanation:

For ax^2 +bx +c, the axis of symmetry is x = -b/(2a). For the offered answer choices, the axes of symmetry are ...

  a) x = -1/(2·2) = -1/4 . . . . . . the function we're looking for

  b) x = -(-1)/(2·2) = 1/4

  c) x = -2/(2·1) = -1

  d) x = -(-2)/(2·1) = 1

The function with axis of symmetry x = -1/4 is f(x) = 2x^2 +x -1.

you can not drow on this

step-by-step explanation: