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

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Answer 1 · 2015-07-17T18:36:48

This is a vertical parabola - sort of U shaped - with vertex at $(5, 0)$ , axis $x = 5$ and intercept with the $y$ axis at $(0, 25)$ .

The vertex is at the minimum value of the function.

$(x-5)^2 >= 0$ since it is the square of a real number.

$(x-5)^2 = 0$ when $x-5 = 0$ , that is when $x=5$ .

Hence the vertex is at $(5, 0)$ .

The axis is vertical, passing through the vertex, so its equation is $x=5$ .

The intercept with the $y$ axis occurs when $x=0$ , so substitute $x=0$ into the equation to find $y = (0 - 5)^2 = (-5)^2 = 25$

So the intercept is at $(0, 25)$

graph{(x-5)^2 [-39.58, 40.42, -5.28, 34.72]}

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