How do you graph $y=(x-3)^2 +5$ ?

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Answer 1 · 2015-08-21T23:14:56

Determine the vertex. Plot the vertex and several points. Draw the parabola as an upward facing curve.

$y=(x-3)^2+5$ is in the vertex form of a parabola, $y=(x-h)^2+k$ , where $h=3$ and $k=5$ . The vertex of this parabola is $(h,k)=(3,5)$ .

Find some points by substituting values for $x$ , making sure to include both positive and negative values.

$x=-2,$ $y=30$
$x=-1,$ $y=21$
$x=0,$ $y=14$
$x=1,$ $y=9$
$x=2,$ $y=6$

Plot the vertex, and then some points. Draw a graph as a curve where the curve turns at the vertex $(3,5)$ .

graph{y=(x-3)^2+5 [-16.82, 15.2, 2.24, 18.25]}

How do you graph y=(x-3)^2 +5y=(x-3)^2 +5?