How do you use the important points to sketch the graph of $y=(x-2)^2$ ?

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Answer 1 · 2018-05-18T02:03:37

See below

$y=(x-2)^2$

$= x^2-4x+4$

Notice that $y$ is a quadratic function of the form $ax^2+bx+c$ , therefore its graph will be a parabola.

$y =0$ at $x=2$ - Notice that this is a coincident root meaning that $y$ will touch the $x-$ axis at $x=2$ and at nowhere else.

Since $a =+1>0, y$ will have a minimum value at
$x=(-b)/(2a) = 4/2 =2$

$:. y_min = 0$ at $x=2$

Hence an "important point" is $(2,0)$ being both the only $x-$ intercept and the minimum of $y$

Next, we can find the $y-$ intercept, where $x=0$

$y(0) = (-2)^2 =4$

So another "important point" is $(0,4)$ the $x-$ intercept

We can see these points on the graph of $y$ below.

graph{(x-2)^2 [-6.83, 10.945, -1.68, 7.21]}

How do you use the important points to sketch the graph of y=(x-2)^2y=(x-2)^2?