Question

For the quadratic function `f(x)= x^2 + 1`, how do you find the y intercept, the equation of the axis of symmetry and the x-coordinate of the vertex?

Answer 1

The y-intercept is (0,1), the equation of the axis of symmetry is x = 0, and the x-coordinate of the vertex is 0.

Explanation:

The vertex form of a parabola is `f(x)=a(x-h)^2+k`. In this, case, `a=1`, `h=0`, and `k=1` giving `f(x)=1(x-0)^2+1` which is the same as `x^2+1`.

The x-coordinate of the y-intercept is always zero, so you find the y-coordinate by finding `f(0)`, in this case `f(0)=(0)^2+1=0+1=1`, so the y-intercept is `(0,1)`.

The equation of the axis of symmetry for a quadratic in vertex form is `x=h`, which in this case is `x=0`. Similarly, the vertex is `(h,k)` which for this quadratic is `(0,1)`, giving the x-coordinate of the vertex as 0. The x-coordinate of the vertex will always be the same as the value for the axis of symmetry.