Question

The graph of which function has an axis of symmetry at x =-1/4 ?

f(x) = 2x2 + x – 1

f(x) = 2x2 – x + 1

f(x) = x2 + 2x – 1

f(x) = x2 – 2x + 1

Answer 1

axis of symmetry is the x value of the vertex

for
y=ax^2+bx+c
x value of vertex=-b/2a

first one
-1/2(2)=-1/4
wow, that is right

answer is first one
f(x)=2x^2+x-1

Answer 2

we know that

The equation of the vertical parabola in vertex form is equal to

`y=a(x-h)^(2)+k`

where

(h,k) is the vertex

The axis of symmetry is equal to the x-coordinate of the vertex

so

`x=h` ------> axis of symmetry of a vertical parabola

we will determine in each case the axis of symmetry to determine the solution

case A)
`f(x)=2x^(2)+x-1`

Convert to vertex form

Group terms that contain the same variable, and move the constant to the opposite side of the equation

`f(x)+1=2x^(2)+x`

Factor the leading coefficient

`f(x)+1=2(x^(2)+0.5x)`

Complete the square. Remember to balance the equation by adding the same constants to each side

`f(x)+1+0.125=2(x^(2)+0.5x+0.0625)`

`f(x)+1.125=2(x^(2)+0.5x+0.0625)`

Rewrite as perfect squares

`f(x)+1.125=2(x+0.25)^(2)`

`f(x)=2(x+0.25)^(2)-1.125`

the vertex is the point
`(-0.25,-1.125)`

the axis of symmetry is

`x=-0.25=-(1)/(4)`

therefore

the function
`f(x)=2x^(2)+x-1` has an axis of symmetry at
`x=-(1)/(4)`

case B)
`f(x)=2x^(2)-x+1`

Convert to vertex form

Group terms that contain the same variable, and move the constant to the opposite side of the equation

`f(x)-1=2x^(2)-x`

Factor the leading coefficient

`f(x)-1=2(x^(2)-0.5x)`

Complete the square. Remember to balance the equation by adding the same constants to each side

`f(x)-1+0.125=2(x^(2)-0.5x+0.0625)`

`f(x)-0.875=2(x^(2)-0.5x+0.0625)`

Rewrite as perfect squares

`f(x)-0.875=2(x-0.25)^(2)`

`f(x)=2(x-0.25)^(2)+0.875`

the vertex is the point
`(0.25,0.875)`

the axis of symmetry is

`x=0.25=(1)/(4)`

therefore

the function
`f(x)=2x^(2)-x+1` does not have a symmetry axis in
`x=-(1)/(4)`

case C)
`f(x)=x^(2)+2x-1`

Convert to vertex form

Group terms that contain the same variable, and move the constant to the opposite side of the equation

`f(x)+1=x^(2)+2x`

Complete the square. Remember to balance the equation by adding the same constants to each side

`f(x)+1+1=x^(2)+2x+1`

`f(x)+2=x^(2)+2x+1`

Rewrite as perfect squares

`f(x)+2=(x+1)^(2)`

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

the vertex is the point
`(-1,-2)`

the axis of symmetry is

`x=-1`

therefore

the function
`f(x)=x^(2)+2x-1` does not have a symmetry axis in
`x=-(1)/(4)`

case D)
`f(x)=x^(2)-2x+1`

Convert to vertex form

Group terms that contain the same variable, and move the constant to the opposite side of the equation

`f(x)-1=x^(2)-2x`

Complete the square. Remember to balance the equation by adding the same constants to each side

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

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

Rewrite as perfect squares

`f(x)=(x-1)^(2)`

the vertex is the point
`(1,0)`

the axis of symmetry is

`x=1`

therefore

the function
`f(x)=x^(2)-2x+1` does not have a symmetry axis in
`x=-(1)/(4)`

the answer is

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