Find the Vertex f(x)=x^2-8x+16

Problem

ƒ(x)=x2−8*x+16

Solution

1. Identify the coefficients of the quadratic function in the standard form ƒ(x)=a*x2+b*x+c

a=1

b=−8

c=16

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

x=(−(−8))/(2*(1))

x=8/2

x=4

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

ƒ(4)=(4)2−8*(4)+16

ƒ(4)=16−32+16

ƒ(4)=0

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

Vertex=(4,0)

Final Answer

Vertex of *ƒ(x)=x2−8*x+16* is *(4,0)

Want more problems? Check here!