Find the Vertex Form f(x)=x^2+x+5/4

$f (x) = x^{2} + x + \frac{5}{4}$

Set the polynomial equal to $y$ to find the properties of the parabola.

$y = x^{2} + x + \frac{5}{4}$

Complete the square for $x^{2} + x + \frac{5}{4}$ .

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Use the form $a x^{2} + b x + c$ , to find the values of $a$ , $b$ , and $c$ .

$a = 1, b = 1, c = \frac{5}{4}$

Consider the vertex form of a parabola.

$a {(x + d)}^{2} + e$

Substitute the values of $a$ and $b$ into the formula $d = \frac{b}{2 a}$ .

$d = \frac{1}{2 (1)}$

Cancel the common factor of $1$ .

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Cancel the common factor.

$d = \frac{1}{2 \cdot 1}$

Rewrite the expression.

$d = \frac{1}{2}$

Find the value of $e$ using the formula $e = c - \frac{b^{2}}{4 a}$ .

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Simplify each term.

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One to any power is one.

$e = \frac{5}{4} - \frac{1}{4 \cdot 1}$

Multiply $4$ by $1$ .

$e = \frac{5}{4} - \frac{1}{4}$

Combine the numerators over the common denominator.

$e = \frac{5 - 1}{4}$

Subtract $1$ from $5$ .

$e = \frac{4}{4}$

Divide $4$ by $4$ .

$e = 1$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $a {(x + d)}^{2} + e$ .

${(x + \frac{1}{2})}^{2} + 1$

Set $y$ equal to the new right side.

$y = {(x + \frac{1}{2})}^{2} + 1$

Find the Vertex Form f(x)=x^2+x+5/4

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