How do you find the vertex and intercepts for #f(x)= -x^2 + 6x + 3#?

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Answer 1 · 2017-01-08T16:27:49

The vertex is #=(3,6)#
The intercepts are #(0,3)#, #(3+2sqrt3,0)# and #(3-2sqrt3,0)#

The vertex is the highest point or lowest point of the equation,

we complete the squares

We rewrite #f(x)# as

#f(x)=-x^2+6x+3#

#=-(x^2-6x)+3#

#=-(x^2-6x+9)+3+9#

#=-(x-3)^2+12#

We compare this to #f(x)=a(x-h)^2+k#

and the vertex is #(h,k)=(3,6)#

The axis of symmetry is #x=3#

To find the intercepts,

first,

Let #x=0#, #=>#, #(y=3)#

second,

Let #y=0#, then

#-(x-3)^2+12=0#

#(x-3)^2=12#

#x-3=+-sqrt12=+-2sqrt3#

#x=3+-2sqrt3#

Therefore,

The points are #(3+2sqrt3,0)# and #(3-2sqrt3,0)#

The points are #(6.464,0)# and #(-0.464,0)#

graph{(y-(-x^2+6x+3))((x-3)^2+(y-12)^2-0.01)(y-1000x+3000)=0 [-18.1, 17.93, -2.36, 15.67]}