# What are the intercepts of y= x^2 -4x + 5 ?

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### 3 Answers

Where the curve of x^2 - 4x + 5 = 0 intercepts the y-axis, x = 0.

=> y = 0 - 0 + 5 = 5

The y-intercept is (0,5)

When the curve intercepts the x-axis, y= 0

=> x^2 - 4x + 5 = 0

Here we see that (-4)^2 < 4*a*c, the curve does not intersect the x-axis.

**The y-intercept is (0, 5).**

y= x^2 - 4x + 5

We need to find y and x intercepts.

y-intercepts are the points such that the graph meets the y-axis.

==> x = 0

==> y= 5

Then y-intercept is ( 0, 5)

Now we will find x-intercepts.

==> y= 0

==> x^2 - 4x + 5 = 0

==> We notice that there are no real roots for the equation.

Then the graph does not intersect with the x-axis.

**Then the only intercept is the point ( 0, 5)**

The intercepts of y= x^2 -4x + 5 have to be determined.

At the point where the curve meets the x-axis, the value of the y-coordinate is 0.

The equation x^2 - 4x + 5 = 0 has to be solved to determine the x-intercepts.

But the equation x^2 - 4x + 5 = 0 does not have any real roots as (-4)^2 - 4*1*5 = -4 which is negative.

The curve does not intersect the x-axis.

At the point where it meets the y-axis, the x coordinate is 0. This gives y = 0 - 0 + 5 = 0

The y intercept of the curve is 5.