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

Aug 11, 2015

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

$f \left(x\right) = {x}^{2} + 3$

Step 1. Your equation is almost in vertex form.

$f \left(x\right) = a {\left(x - h\right)}^{2} + k$.

Re-write it slightly to get

$f \left(x\right) = 1 {\left(x - 0\right)}^{2} + 3$

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 $y$.

$f \left(0\right) = {0}^{2} + 3 = 0 + 3 = 3$

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

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

Set $f \left(x\right) = 0$ 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 $x = 1$.

$f \left(- 1\right) = {\left(- 1\right)}^{2} + 3 = 1 + 3 = 4$

$f \left(1\right) = {1}^{2} + 3 = 1 + 3 = 4$

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.