Question

How do you graph `f(x)=x^2+3`?

Answer 1

Calculate the vertex and the `x`- and `y`-intercepts and then sketch the graph.

`f(x)= x^2+3`

Step 1. Your equation is almost in vertex form.

`f(x) = a(x-h)^2 +k`.

Re-write it slightly to get

`f(x)= 1(x-0)^2+3`

We see that `a=1`, `h=0`, and `k=3`.

Step 2. Find the vertex.

The vertex is at (`h,k`) or (`0,3`).

Step 3. Find the `y`-intercept.

Set `x=0` and solve for `y`.

`f(0)= 0^2+3 =0+3=3`

The `y`-intercept is at (`0,3`).

Step 4. Find the `x`-intercept(s).

Set `f(x)=0` and solve for `x`.

`0= x^2+3`

`x^2=-3`

`x=±sqrt(-3)= =±isqrt3`

There are no `x`-intercepts.

Step 5. Calculate a few more points.

Try `x=-1` and `x=1`.

`f(-1) = (-1)^2+3 = 1+3=4`

`f(1) = 1^2+3 = 1+3=4`

Step 6. Draw your axes and plot the three points.

Step 6. Draw a smooth parabola that passes through the four points.

And you have your graph.