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 https://brainly.com/question/2491745
2. A problem on function https://brainly.com/question/9590016
3. A problem on axis of symmetry https://brainly.com/question/1286775
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.
