Mathematics, 30.08.2019 07:30 Undrea43

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

Answers: 2

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Answer from: glitch14

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

f(x) = 2x² + x – 1Further explanation

Discriminant of quadratic equation ( ax² + bx + c = 0 ) could be calculated by using :

D = b² - 4 a c

From the value of Discriminant , we know how many solutions the equation has by condition :

D < 0 → No Real Roots

D = 0 → One Real Root

D > 0 → Two Real Roots

Let us now tackle the problem!

An axis of symmetry of quadratic equation y = ax² + bx + c is :

$\large {\boxed {x = \frac{-b}{2a} } }$

Option 1 :

f(x) = 2x² + x – 1 → a = 2 , b = 1 , c = -1

Axis of symmetry → $x = \frac{-b}{2a} = \frac{-1}{2(2)} = -\frac{1}{4}$

Option 2 :

f(x) = 2x² – x + 1 → a = 2 , b = -1 , c = 1

Axis of symmetry → $x = \frac{-b}{2a} = \frac{-(-1)}{2(2)} = \frac{1}{4}$

Option 3 :

f(x) = x² + 2x – 1 → a = 1 , b = 2 , c = -1

Axis of symmetry → $x = \frac{-b}{2a} = \frac{-2}{2(1)} = -1$

Option 4 :

f(x) = x² – 2x + 1 → a = 1 , b = -2 , c = 1

Axis of symmetry → $x = \frac{-b}{2a} = \frac{-(-2)}{2(1)} = 1$

Learn moreSolving Quadratic Equations by Factoring : Determine the Discriminant : Formula of Quadratic Equations : Answer details

Grade: High School

Subject: Mathematics

Chapter: Quadratic Equations

Keywords: Quadratic , Equation , Discriminant , Real , Number

The graph of which function has an axis of symmetry at x =-1/4 ? f(x) = 2x2 + x – 1 f(x) = 2x2 – x

Answer from: IntellTanito

The graph of function $\boxed{f(x)=2x^{2}+x-1}$ has an axis of symmetry as $\boxed{x=-\frac{1}{4}}$ .

Further explanation:

The standard form of a quadratic equation is as follows:

$\boxed{f(x)=ax^{2}+bx+c}$

The vertex form of a quadratic equation is as follows:

$\boxed{g(x)=a(x-h)^{2}+k}$

Axis of symmetry is the line which divides the graph of the parabola in two perfect halves.

The formula for axis of symmetry of a quadratic function is given as follows:

$\boxed{x=-\dfrac{b}{2a}}$

The first function is given as follows:

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

The above function is in standard form with a=2 , b=1 and c=-1 .

Then its axis of symmetry is calculated as,

$\begin{aligned}x&=-\dfrac{b}{2a}\\&=-\dfrac{1}{2\times2}\\&=-\dfrac{1}{4}\end{aligned}$

The axis of symmetry of first function is $x=-\frac{1}{4}$ .

Express the function $f(x)=2x^{2}+x-1$ in its vertex form,

$\begin{aligned}f(x)&=2x^{2}+x-1\\&=(\sqrt{2}x)^{2}+\left(2\times \sqrt{2}x\times \dfrac{1}{2\sqrt{2}}\right)-1+\left(\dfrac{1}{2\sqrt{2}}\right)^{2}-\left(\dfrac{1}{\sqrt{2}}\right)^{2}\\&=\left(\sqrt{2}x+\dfrac{1}{2\sqrt{2}}\right)^{2}-1-\dfrac{1}{8}\\&=\left[\sqrt{2}\left(x+\dfrac{1}{4}\right)\right]^{2}-\dfrac{9}{8}\\&=2\left(x-\left(-\dfrac{1}{4}\right)\right)^{2}-\dfrac{9}{8}\end{aligned}$

The above equation is in the vertex form with a=2 , $h=-\dfrac{1}{4}$ and $k=-\dfrac{9}{8}$ .

Therefore, its axis of symmetry is given as,

$\begin{aligned}x&=h\\x&=-\dfrac{1}{4}\end{aligned}$

The graph of function $f(x)=2x^{2}+x-1$ is shown in Figure 1.

The second function is given as follows:

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

The above function is in standard form with a=2 , b=-1 and c=1 .

Then its axis of symmetry is calculated as,

$\begin{aligned}x&=-\dfrac{b}{2a}\\&=-\dfrac{(-1)}{2\times2}\\&=\dfrac{1}{4}\end{aligned}$

The axis of symmetry of second function is $x=\frac{1}{4}$ .

The third function is given as follows:

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

The above function is in standard form with a=1 , b=2 and c=-1 .

Then its axis of symmetry is calculated as,

$\begin{aligned}x&=-\dfrac{b}{2a}\\&=-\dfrac{2}{2\times1}\\&=-1\end{aligned}$

The axis of symmetry of third function is x=-1 .

The fourth function is given as follows:

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

The above function is in standard form with a=1 , b=-2 and c=1 .

Then its axis of symmetry is calculated as,

$\begin{aligned}x&=-\dfrac{b}{2a}\\&=-\dfrac{-2}{2\times1}\\&=1\end{aligned}$

The axis of symmetry of fourth function is x=1 .

Therefore, the function $\boxed{f(x)=2x^{2}+x-1}$ has an axis of symmetry as $\boxed{x=-\frac{1}{4}}$ .

Learn more:

1. A problem on graph

2. A problem on function

3. A problem on axis of symmetry

Answer details:

Grade: High school

Subject: Mathematics

Chapter: Functions

Keywords:Graph, function, axis, f(x), 2x^2+x-1, axis of symmetry, symmetry, vertex, perfect halves, graph of a function, x =- 1/4.

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The graph of which function has an axis of symmetry at x =-1/4 ?
f(x) = 2x2 + x – 1
f(x...