Find the Vertex f(x)=x^2-4x-12

Problem

ƒ(x)=x^2−4*x−12

Solution

1. Identify the coefficients of the quadratic function in the form ƒ(x)=a*x^2+b*x+c

a=1

b=−4

c=−12

2. Calculate the xcoordinate of the vertex using the formula x=(−b)/(2*a)

x=(−(−4))/(2*(1))

x=4/2

x=2

3. Substitute the xvalue back into the original function to find the ycoordinate.

ƒ(2)=(2)^2−4*(2)−12

ƒ(2)=4−8−12

ƒ(2)=−16

4. State the vertex as the ordered pair (x,y)

Vertex=(2,−16)

Final Answer

Vertex of *ƒ(x)=x^2−4*x−12* is *(2,−16)

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