find the vertex of the parabola

Question

f(x)=x^2-4x-5

 

The vertex of the given parabola is _____.  (typed an ordered pair)

Answer 1

2 other ways to do this without using derivatives directly:

 

1)

Rewrite in parabolic form by completing the square.

 

f(x) = x²-4x-5   = (x² - 4x + 4 ) - 5 - 4    (adding 4 inside means to subtract 4 outside to keep the net change = 0.

 

f(x) = (x-2)² - 9

 

f(x) = a(x-h)² = k... (h, k) is the vertex.

 

Answer:  (2, -9)

 

2)

 

-b/(2a) is the x coordinate of the vertex.   f(-b/(2a) = y coordinate.

 

a = 1, b = -4, c= -5

x = -b/(2a) = -(-4)/(2)(1) = 4/2 = 2

 

y = f(2) = (2)² -4(2) - 5 = 4 - 8 - 5 = -9

 

Answer:  (2, -9)

 

 

 

 

Answer 2

I don't know if you know about derivatives yet, but the first derivative of f(x) with respect to x is

 

[df(x)/dx] = 2x -4

 

Setting this equal to 0 gives x = 2 which is the x-coordinate of the vertex.

 

Solving f(x) at x = 2 gives (2)² - (4)(2) - 5 = -9

 

The vertex of the parabola is at the point (2, -9).