Question

How to graph f(x)=(x-2)^2+5

Answer 1

Draw y = x² and shift it 2 units right and 5 units up.

$f(x)=(x-\boxed{2})^2+\boxed{5}$

$y=x^2\\\\for\ x=\pm2\to y=(\pm2)^2=4\to(-2;\ 4);\ (2;\ 4)\\\\for\ x=\pm1\to y=(\pm1)^2=1\to(-1;\ 1);\ (1;\ 1)\\\\for\ x=0\to y=0^2=0\to(0;\ 0)$

Look at the picture.

Answer 2

That's a quadratic, a nice parabola in vertex form.

The parabola has a positive x^2 term, so it's a CUP, concave up positive. It will have a minimum at the vertex, which is (2,5). Plot that point.

Now we need a couple of guide points to draw the usual parabola going up from both sides of its vertex. We try x=0 giving (0,9) and see that x=4 also gives 9, (4,9). Plot the parabola through those two points and the vertex and you're done.