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DeanR

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.



gmany

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

[tex]f(x)=(x-\boxed{2})^2+\boxed{5}[/tex]


[tex] 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) [/tex]

Look at the picture.

Ver imagen gmany

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