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? (0, 1) (1, 3) (1, 7) (2, 1)?????

Answer 1

Hello :
-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)  

Answer 2

Answer:

Option B is correct.

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

Explanation:

Firstly, convert the given equation into vertex form:

$y =m(x-a)^2+b$          .....,[1]; where m is the factor related to the horizontal spread of the parabola and (a,b) is the (x,y) coordinate of vertex.

Given:

y =f(x) = $-2x^2+4x+1$  and the axis of symmetry, x =1

y = $-2(x^2-2x)+1$ or

y = $-2(x^2-2x+1-1)+1$ or

y = $-2(x^2-2x+1)+2+1$ or

y = $-2(x-1)^2+3$         Using identity: $(a-b)^2 = a^2-2ab+b^2$

On comparing above equation with equation [1] we get;

m = -2 , a = 1 and b = 3

Therefore, the vertex of the function located at ( 1, 3).