Q. The graph of which function has an axis of symmetry at \(x = \frac{1}{4}\)?

Q: How do you find the axis of symmetry for a parabola ax^2+bx+c?

Use the formula x = -\dfrac{b}{2a}. That gives the vertical line through the vertex; it depends only on a and b.

Q: Which of the given functions has axis x=\tfrac{1}{4}?.

f(x)=2x^2-x+1. Here a=2,b=-1, so x=-\dfrac{-1}{2\cdot2}=\tfrac{1}{4}.

Q: What is the axis of symmetry for each listed function?

2x^2+x-1:\ x=-\tfrac{1}{4}. 2x^2-x+1:\ x=\tfrac{1}{4}. x^2+2x-1:\ x=-1. x^2-2x+1:\ x=1.

Q: How do you find the vertex coordinates?

Vertex x-coordinate x_v=-\dfrac{b}{2a}. Vertex y-coordinate y_v=f(x_v); plug x_v into the function or use completion of the square.

Q: How does completing the square give the axis?

Rewrite ax^2+bx+c=a\bigl(x+\tfrac{b}{2a}\bigr)^2-\tfrac{b^2}{4a}+c. The squared term centers at x=-\tfrac{b}{2a}, so axis is x=-\tfrac{b}{2a}.

Q: Does the constant term c affect the axis?

Does the constant term c affect the axis?

Q: If a parabola has axis x=\tfrac{1}{4}, what relation must a and b satisfy?

Solve -\dfrac{b}{2a}=\tfrac{1}{4}. This gives b=-\dfrac{a}{2}.

Answer

Axis at \( x = \dfrac{-b}{2a} \). Set \( \dfrac{-b}{2a} = \dfrac{1}{4} \Rightarrow b = \dfrac{-a}{2} \). For \( f(x) = 2x^2 – x + 1 \), \( a = 2 \), \( b = -1 \) and \( -1 = \dfrac{-2}{2} \), so the correct function is \( f(x) = 2x^2 – x + 1 \).

Detailed Explanation

Recall the formula for the axis of symmetry of a parabola given by a quadratic function f(x) = ax2 + bx + c:\[ x = -\frac{b}{2a} \]This is the x-coordinate of the vertex and thus the axis of symmetry.
Set this equal to one-quarter and solve for the relationship between a and b:\[ -\frac{b}{2a} = \frac{1}{4} \]Multiply both sides by 2a:
\[ -b = \frac{2a}{4} = \frac{a}{2} \]

So

\[ b = -\frac{a}{2} \]

Any quadratic whose coefficients a and b satisfy b = −a/2 will have axis x = 1/4.
Check each given function (identify a and b, then compute the axis x = −b/(2a)):
1. f(x) = 2×2 + x − 1: a = 2, b = 1.\[ x = -\frac{b}{2a} = -\frac{1}{2\cdot 2} = -\frac{1}{4} \]Not equal to 1/4.
2. f(x) = 2×2 − x + 1: a = 2, b = −1.\[ x = -\frac{b}{2a} = -\frac{-1}{2\cdot 2} = \frac{1}{4} \]Equals 1/4. This one satisfies the condition.
3. f(x) = x2 + 2x − 1: a = 1, b = 2.\[ x = -\frac{b}{2a} = -\frac{2}{2\cdot 1} = -1 \]Not equal to 1/4.
4. f(x) = x2 − 2x + 1: a = 1, b = −2.\[ x = -\frac{b}{2a} = -\frac{-2}{2\cdot 1} = 1 \]Not equal to 1/4.
Conclusion: The graph with axis of symmetry at x = one-quarter is\[ f(x) = 2x^{2} – x + 1 \]

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Algebra FAQs

How do you find the axis of symmetry for a parabola \(ax^2+bx+c\)?

Use the formula \(x = -\dfrac{b}{2a}\). That gives the vertical line through the vertex; it depends only on \(a\) and \(b\).

Which of the given functions has axis \(x=\tfrac{1}{4}\)?.

\(f(x)=2x^2-x+1\). Here \(a=2,b=-1\), so \(x=-\dfrac{-1}{2\cdot2}=\tfrac{1}{4}\).

What is the axis of symmetry for each listed function?

\(2x^2+x-1:\ x=-\tfrac{1}{4}.\) \(2x^2-x+1:\ x=\tfrac{1}{4}.\) \(x^2+2x-1:\ x=-1.\) \(x^2-2x+1:\ x=1.\)

How do you find the vertex coordinates?

Vertex x-coordinate \(x_v=-\dfrac{b}{2a}\). Vertex y-coordinate \(y_v=f(x_v)\); plug \(x_v\) into the function or use completion of the square.

How does completing the square give the axis?

Rewrite \(ax^2+bx+c=a\bigl(x+\tfrac{b}{2a}\bigr)^2-\tfrac{b^2}{4a}+c\). The squared term centers at \(x=-\tfrac{b}{2a}\), so axis is \(x=-\tfrac{b}{2a}\).

Does the constant term \(c\) affect the axis?

If a parabola has axis \(x=\tfrac{1}{4}\), what relation must \(a\) and \(b\) satisfy?

Solve \( -\dfrac{b}{2a}=\tfrac{1}{4}\). This gives \( b=-\dfrac{a}{2}\).

Can more than one quadratic have the same axis?

Yes. Any quadratics with coefficients satisfying the same ratio \(b/(2a)\) share the same axis; among the four given, only one meets \(x=\tfrac{1}{4}\).

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