What is the vertex of the parabola #y=-x^2-2x+3#?
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There is a lovely and straightforward (which makes it all the lovelier) rule for working out vertices such as this one.
Think of the general parabola: #y=ax^2+bx+c#, where #a!=0#
The formula for finding the #x#-vertex is #(-b)/(2a)# and to find the #y#-vertex, you insert the value you found for #x# into the formula.
Using your question #y=-x^2-2x+3# we can establish the values of #a, b, #and #c#.
In this case:
#a=-1#
#b=-2#; and
#c=3#.
To find the #x#-vertex we need to replace the values for #a# and #b# in the formula given above (#color(red)((-b)/(2a))#):
#=(-(-2))/(2*(-1))=2/(-2)=-1#
So we now know that the #x#-vertex is at #-1#.
To find the #y#-vertex, go back to the original question and replace all the instances of #x# with #-1#:
#y=-x^2-2x+3#
#y=-(-1)^2-2*(-1)+3#
#y=-1+2+3#
#y=4#
We now know that the #x#-vertex is at #-1# and the #y#-vertex is at #4# and this can be written in coordinate format:
#(-1,4)#
