Question

The axis of symmetry for the function f(x) = –2x2 + 4x + 1 is the line x = 1. Where is the vertex of the function located?

Answer 1

Answer:

(1;3)

Step-by-step explanation:

1st way for solution:

-2x²+4x+1= -2(x-1)²+b

find : b

-2x²+4x+1 = -2(x²-2x+1)+b

-2x²+4x+1 = -2x² +4x -2+b

-2+b = 1

b = 3

-2x²+4x+1= -2(x-1)²+3

the vertex of the function located is : (1, 3)

2nd way:

x =1

f(1)= -2+4+1=3

So answer is (1;3)

3rd way:

X is given, this is formula for y

n= (4ac-b^2)/4a=(4*(-2)*1-4^2)/4*(-2)=-24/-8=3

So the vertex is (1;3)

Answer 2

The vertex of the function is located at (1,3)

How to determine the vertex?

The function is given as:

f(x) = -2x^2 + 4x + 1

Differentiate

f'(x) = -4x + 4

Set to 0

-4x + 4 = 0

Subtract 4 from both sides

-4x =- 4

Divide by -4

x = 1

Substitute 1 for x in f(x)

f(1) = -2(1)^2 + 4(1) + 1

Evaluate

f(1) = 3

Hence, the vertex of the function is located at (1,3)

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