Question 596732
The function {{{f(x)=(1/2)x^2+1}}} takes the same value for x as for -x.
{{{f(-x)=f(x)}}}
Functions that do that are called even functions and are symmetrical with respect to the y-axis.
The y-axis is the line {{{x=0}}}.
You can get five points if you just calculate the value of the function at x=0, and at a couple of positive x values. You get the symmetrical points on the side with negative x for free.
For x=0, {{{y=(1/2)0^2+1=0+1=1}}}, which gives you the point (0,1).
For x=2, {{{y=(1/2)2^2+1=4/2+1=2+1=3}}}, which gives you the point (2,3).
For x=4, {{{y=(1/2)4^2+1=16/2+1=8+1=9}}}, which gives you the point (4,9).
No calculations are required for the symmetrical points (-2,3) and (-4,9); you just change the sign of the x coordinate.
Next, plot the five points {{{drawing(300,300,-5,5,-1,11,
grid(1),
red(circle(0,1,0.2)),
red(circle(2,3,0.2)),
red(circle(-2,3,0.2)),
red(circle(4,9,0.2)),
red(circle(-4,9,0.2))
)}}} and then draw a reasonable line
{{{graph(300,300,-5,5,-1,11,x^2/2+1)}}}
Doing it with pencil and paper, show my calculations first, with few, if any, words. Then I would have the line with the points clearly marked on just one graph, and I would include a table with the x and y values next to the graph. However, you should do it in whatever style your teacher prefers.