I want to sketch the graph of this function:
$y = x^2 + 2$
Since the coefficient of $x^2$ is $1$, the graph opens upwards.
When $y = 0$, I find that the graph does not cut the $x$-axis. (It is above it.)
When $x = 0$, I find that the graph cuts the $y$-axis at $(0, 2)$.
Hence, the equation of the line of symmetry is $x = 0$.
I still do not know the steepness of the parabola.
From this limited information, how do I sketch the graph of this function without using graphing software?
What additional formula(s) or method(s) would I need to sketch this graph? A demonstration based on my problem would be helpful.
Best Answer
Your observations are correct thus far. You could include several more points in your graph, such as $x=1,2,4,\ldots$. This would already give you a good idea of the ''steepness'' of the function.
Then you could study the derivative of your function, to find its slopes at various points: $$ (x^2+2)' = 2x, $$ so you already know that at $x=\frac12$ you have a slope of $1$ and at $x=-\frac12$ a slope of $-1$. Then find the two $x$'s where the slope is twice that, and so forth.
This should already be more than enough to sketch a graph of the function.