Question

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

Explanation:

Step1: Recall axis of symmetry formula

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

Step2: Analyze \( f(x) = 2x^2 + x - 1 \)

Here, \( a = 2 \), \( b = 1 \).
Substitute into the formula: \( x = -\frac{1}{2(2)} = -\frac{1}{4} \).

Step3: Analyze \( f(x) = x^2 - 2x + 1 \)

Here, \( a = 1 \), \( b = -2 \).
Substitute: \( x = -\frac{-2}{2(1)} = 1 \).

Step4: Analyze \( f(x) = x^2 + 2x - 1 \)

Here, \( a = 1 \), \( b = 2 \).
Substitute: \( x = -\frac{2}{2(1)} = -1 \).

Step5: Analyze \( f(x) = 2x^2 - x + 1 \)

Here, \( a = 2 \), \( b = -1 \).
Substitute: \( x = -\frac{-1}{2(2)} = \frac{1}{4} \).

Answer:

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