How do you find the vertex and the intercepts for #y=-x^2+x+12#?

The vertex form of a parabola is
#color(white)("XXX")y=color(m)(x-color(blue)(a))+color(red)(b)#
with vertex at #(color(blue)(a),color(red)(b))#

Converting the given #y=-x^2+x+12# into vertex form:

#color(white)("XXX")y=color(green)((-1))(x^2-x+(1/2)^2)+12 + (1/2)^2#

#color(white)("XXX")y=color(green)((-1))(x-color(blue)(1/2))^2+color(red)(49/4)#

giving us the vertex at #(color(blue)(1/2,)color(red)(49/4))#

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The y-intercept is the value of #y# when #color(magenta)(x=0)#

#y=-x^2+x+12# when #color(magenta)(x=0)#
#rarr y=-color(magenta)(0)^2+color(magenta)(0)+12=12#

So the y-intercept is #0#

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The x-intercepts are the values of #x# when #color(brown)(y=0)#

#y=-x^2+x+12# when #color(brown)(y=0)#
#rarr -x^2+x+12=color(brown)(0) color(white)("XX")orcolor(white)("XX")x^2-x-12=0#

This can be factored as
#color(white)("XXX")(x-4)(x+3)=0#
#rarr x=4color(white)("XX")orcolor(white)("XX")x=-3#

So the x-intercepts are #{4,-3}#

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Here is the graph of this relation for verification purposes:
graph{-x^2+x+12 [-10.29, 18.19, -1.36, 12.9]}