I need help to find the x-intercepts y=x^2-2x+4?

Question

can you show me the steps?

Answer 1

x² - 2x + 4 = 0 for x-intercepts
x = [ 2 ± √(4 - 16) ] / 2
x = [ 2 ± √(- 12) ] / 2
x = [2 ± i √(12)] / 2
x = [ 2 ± 2i √3 ] / 2
x = 1 ± i √3

Answer 2

Use the quadratic formula.

x= [-b +- sqrt (b^2 - 4ac) ] / (2a)

=2 +- sqrt ((-2)^2 - 4(1)(4)) / (2(1))

When you use it, you'll find that you will be taking the square root of a negative number. You can then conclude that no roots (x intercepts) exist. The vertex of this parabola lies above the x-axis.

You can also check with the discriminant:

if b^2 - 4ac > 0 : There are two real roots

b^2 - 4ac = 0 : There is one root

b^2 - 4ac < 0 : There are no real roots