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What is the vertex of #f(x)= -x^2 + 6x + 3#?

Answer 1

#(3, 12)#

Use #x_(vertex)=(-b)/(2a)# In this case, #a=-1, b=6#, so #x_(vertex)=3# Then, the coordinate is #(3, f(3)) = (3, 12)#

Source of this formula:

We know the vertex's x position is the average of the two solutions. To find the x component of the vertex, we take the average: #x_(vertex)=(x_1 + x_2) / 2# We also know that: #x_(1, 2)=(-b+-sqrt(b^2-4ac))/(2a)=(-b+-sqrt(Delta))/(2a)# where #Delta# is the discriminate.

So then we can derive that: #x_(vertex)=1/2 ((-b+sqrt(Delta))/(2a) + (-b-sqrt(Delta))/(2a)) =1/2((-b + sqrt(Delta) + -b - sqrt(Delta)) / (2a)) =1/2((-2b)/(2a))# #=(-b)/(2a)#

Voila.

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Answer 2

Nathan Axel

To find the vertex of the quadratic function f(x) = -x^2 + 6x + 3, use the formula for the x-coordinate of the vertex: x = -b/2a, where a = -1 and b = 6. Substitute these values into the formula to find the x-coordinate of the vertex. Then, substitute the x-coordinate back into the original function to find the corresponding y-coordinate. Therefore, the vertex of the function is (3, 12).

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Answer from HIX Tutor

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