Question

Which function in vertex form is equivalent to f(x) = x² + x +1?

Answer 1

`f(x)=\left(x+(1)/(2)\right)^2+(3)/(4)`

Explanation:

`\boxed{\begin{minipage}{5.6 cm}\underline{Vertex form of a quadratic equation}\\\\$y=a(x-h)^2+k$\\\\where:\\ \phantom{ww}$\bullet$ $(h,k)$ is the vertex. \\ \phantom{ww}$\bullet$ $a$ is some constant.\\\end{minipage}}`

To find the vertex form of the given function, complete the square.

Given function:

`f(x)=x^2+x+1`

Add and subtract the square of half the coefficient of the term in x:

`\implies f(x)=x^2+x+\left((1)/(2)\right)^2+1-\left((1)/(2)\right)^2`

`\implies f(x)=x^2+x+(1)/(4)+1-(1)/(4)`

`\implies f(x)=x^2+x+(1)/(4)+(3)/(4)`

Factor the perfect square trinomial formed by the first three terms:

`\implies f(x)=\left(x+(1)/(2)\right)^2+(3)/(4)`

Answer 2

• f(x) = (x + 1/2)² + 3/4

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Vertex form of quadratic equation:

• f(x) = a(x - h)² + k

Convert the given expression into vertex form by completing the square:

• f(x) =
• x² + x + 1 =
• x² + 2*(1/2)x + (1/2)² + 1 - (1/2)² =
• (x + 1/2)² + 1 - 1/4 =
• (x + 1/2)² + 3/4