Question 13 Rewrite x^2-7x+13 in the form (x-h)^2+k by completing the square. (x+(7)/(2))^2+(3)/(4) (x-(7)/(2))^2-(3)/(4) (x-(7)/(2))^2+(3)/(4) (x+(7)/(2))^2-(3)/(4)

## Step 1: Isolate the $x$ terms<br />### Group the terms with $x$ together: $(x^2 - 7x) + 13$<br /><br />## Step 2: Complete the square<br />### Take half of the coefficient of the $x$ term ($-7$), square it ($(\frac{-7}{2})^2 = \frac{49}{4}$), and add and subtract it inside the parenthesis: $(x^2 - 7x + \frac{49}{4} - \frac{49}{4}) + 13$<br /><br />## Step 3: Simplify the perfect square<br />### Factor the perfect square trinomial: $(x - \frac{7}{2})^2 - \frac{49}{4} + 13$<br /><br />## Step 4: Combine the constants<br />### Combine the constant terms: $(x - \frac{7}{2})^2 + \frac{52 - 49}{4} = (x - \frac{7}{2})^2 + \frac{3}{4}$