How do you graph $f(x)=(x-6)^2+3$ and identify the x intercepts, vertex?

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Answer 1 · 2017-09-27T21:27:48

See below

The quadratic is already in vertex form:

$f(x)=y=a(x-h)^2+k$

So, you can identify the vertex by inspection:

$(h,k)=(6,3)$

To find the $x$ -intercepts, set $y=0$ :

$(x-6)^2+3=0$

$(x-6)^2=-3$

$x-6=sqrt(-3)$

$x=6+-sqrt(-3)$

$x=6+3i$
$x=6-3i$

As you can see, the $x$ -intercepts are imaginary.

graph{(x-6)^2+3 [-5, 10, -2, 8]}

How do you graph f(x)=(x-6)^2+3f(x)=(x-6)^2+3 and identify the x intercepts, vertex?