Question

How do you find the vertex and axis of symmetry, and then graph the parabola given by: `y = x^2 - 4x + 1`?

Answer 1

`aos = 2`
`vertex = (2, -3)`
`"y-int" = 1`
`"x-int" = 2-sqrt3 and 2+sqrt3`

Explanation:

Standard form is:

`f(x)=ax^2+bx+c`

your equations is already in standard form.

`a=1`
`b= -4`
`c =1`

The formula for the axis of symmetry is:

`aos = (-b)/(2a)`

`aos = (-(-4))/(2*1)`

`aos = 2`

Formula for the vertex point `(x,y)` in:

`vertex = (aos, f(aos))`

what is f(aos)? that means you put the value you found for the aos back into the function as x and solve for y:

`y= x^2-4x+1`

`y = 2^2-4(2) +1`

`y = -3`

so the vertex is: `(2, -3)`

Finally to help you graph the y-intercept is `c` so:

y-int = 1

also the x-intercepts are the roots if you factor the polynomial, you would need to use the quadratic formula to factor this one and the roots will be:

`x = 2+sqrt3`
and
`x = 2-sqrt3`