Question

If function g(x) = x2 + 2 is a transformation of the quadratic parent function, f(x) = x2, where is the y-intercept of function g located?

Answer 1

We have been given that function $g(x)=x^2+2$ is a transformation of the quadratic parent function $f(x)=x^2$. We are asked to find the y-intercept of function g.

We know that the function $f(x)=x^2$ is an upward opening parabola with vertex at point (0,0).

We know that vertex form of a parabola is in form $y=a(x-h)^2+k$, where point (h,k) represents vertex of parabola.

We can rewrite g(x) as:

$g(x)=(x-0)^2+2$

The vertex of the function g(x) is at point (0,2).

We know that the vertex of a function is the point, when x is equal to 0. Therefore, the y-intercept of the g is at (0,2).

Answer 2

Answer:

Step-by-step explanation:

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