Question

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

Answer 1

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

$x=-0.25=-\frac{1}{4}$

therefore

the function $f(x)=2x^{2}+x-1$ has an axis of symmetry at $x=-\frac{1}{4}$

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

$x=0.25=\frac{1}{4}$

therefore

the function $f(x)=2x^{2}-x+1$ does not have a symmetry axis in $x=-\frac{1}{4}$

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 $x=-\frac{1}{4}$  

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 $x=-\frac{1}{4}$

the answer is

$f(x)=2x^{2}+x-1$

Answer 2

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