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

##### 1 Answer
Mar 29, 2018

x-intercept = (4±2sqrt(2))/4

y-intercept = 1

#### Explanation:

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

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

$a x + b x + c = 0$ ($2 {x}^{2} - 4 x + 1$)

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±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

$f \left(x\right) = 2 {\left(0\right)}^{2} - 4 \left(0\right) + 1$

$f \left(x\right) = 2 \left(0\right) - 0 + 1$

$f \left(x\right) = 0 - 0 + 1$

#f(x)=1