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

Explanation:

Step1: Recall axis - of - symmetry formula

For a quadratic function $f(x)=ax^{2}+bx + c$, the axis of symmetry is given by $x =-\frac{b}{2a}$.

Step2: Check option 1

For $f(x)=2x^{2}+x - 1$, where $a = 2$ and $b = 1$. Then $x=-\frac{b}{2a}=-\frac{1}{2\times2}=-\frac{1}{4}$.

Step3: Check option 2

For $f(x)=2x^{2}-x + 1$, where $a = 2$ and $b=-1$. Then $x =-\frac{b}{2a}=-\frac{-1}{2\times2}=\frac{1}{4}$.

Step4: Check option 3

For $f(x)=x^{2}+2x - 1$, where $a = 1$ and $b = 2$. Then $x=-\frac{b}{2a}=-\frac{2}{2\times1}=-1$.

Step5: Check option 4

For $f(x)=x^{2}-2x + 1$, where $a = 1$ and $b=-2$. Then $x=-\frac{b}{2a}=-\frac{-2}{2\times1}=1$.

Answer:

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