How do you find the x-int, y-int and graph $f(x)= 2x^2 - 4x +1$ ?

Question

5029 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-29T22:16:23

x-intercept $= (4±2sqrt(2))/4$

y-intercept = 1

To find the x-intercept, let f(x), which is y equal to zero:

$0=2x^2-4x+1$ (Since this is not factorable, you would use the quadratic formula to solve for x)

$ax+bx+c=0$ ( $2x^2-4x+1$ )

$x= (-b±sqrt(b^2 -4ac))/(2a)$

$x= (-(-4)±sqrt((-4)^2 -4(2)(1)))/(2(2))$

$x= (4±sqrt(16 -8))/4$

$x= (4±sqrt(8))/4$

$x= (4±2sqrt(2))/4$

Now to find the y-intercept, make x in the equation = to zero:

$f(x)=2x^2−4x+1$

$f(x) = 2(0)^2-4(0)+1$

$f(x)=2(0)-0+1$

$f(x)=0-0+1$

#f(x)=1

How do you find the x-int, y-int and graph f(x)= 2x^2 - 4x +1f(x)= 2x^2 - 4x +1?