f(x)=x^2-4x+6

Question

I need help in finding the y-intercept

equation of the-axis of symmetry 

and the vertex.

 

appreciate the help!

Thank you!

Answer 1

You need to complete the square:

 

f(x) = x² - 4x + 6 = (x² - 4x + 4) + 2 = (x - 2)² + 2

 

The y-intercept is a point of the form (0, f(0)):

     y = f(0) = 6

 

The axis of symmetry is where x-2 = 0.

     x = 2

 

Also, x = 2 is the x-coordinate of the vertex. So the vertex is at (x, y) = (2, f(2))

     (2, 2)

Answer 2

Compare with quadratic equation where 

a=1, b = -4 , c=6

 

x coordinate of vertex is = -b/2a

= 4/2= 2

 

Now plug in x=2 in the original equation to get y coordinate,

y = 4-8+6 = 2

So vertex is (2,2).

Y Intercept is where x = 0, plug in for in original equation, we get

 

Y = 6, is the Y intercept.