How do you find the y intercept, axis of symmetry and the vertex to graph the function $f (x) = x^{2} + 4$ $f(x)=x^2+4$ ?

Question

1497 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-05-15T08:59:23

$y$ $y$ intercept is $4$ $4$ , axis of symmetry is $x = 0$ $x=0$ i.e. $y$ $y$ -axis and vertex is $(0, 4)$ $(0,4)$ .

This equation is of the form

$y = f (x) = a {(x - h)}^{2} + k$ $y=f(x)=a(x-h)^2+k$

whose axis of symmetry is $x - h = 0$ $x-h=0$ and vertex is $(h, k)$ $(h,k)$

As $y = f (x) = x^{2} + 4 = {(x - 0)}^{2} + 4$ $y=f(x)=x^2+4=(x-0)^2+4$

Hence axis of symmetry is $x = 0$ $x=0$ i.e. $y$ $y$ -axis

Vertex is $(0, 4)$ $(0,4)$

To find $y$ $y$ intercept we should have $x = 0$ $x=0$ and at $x = 0$ $x=0$ ,

we have $y = 0^{2} + 4 = 4$ $y=0^2+4=4$

Hence $y$ $y$ intercept is $4$ $4$ .

graph{x^2+4 [-10, 10, 0, 10]}

How do you find the y intercept, axis of symmetry and the vertex to graph the function f(x)=x2+4f(x)=x^2+4?