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

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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