Question

Rewrite the function by completing the square. f(x)= 2 x^{2} +13 x +20f(x)=2x 2 +13x+20f, left parenthesis, x, right parenthesis, equals, 2, x, squared, plus, 13, x, plus, 20 f(x)=f(x)=f, left parenthesis, x, right parenthesis, equals (x+(x+left parenthesis, x, plus )^2+) 2 +right parenthesis, squared, plus

Answer 1

Answer:

• f(x) = 2(x + 13/4)² - 9/8

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Given

• Function f(x) = 2x² + 13x + 20

Rewrite it by completing the square

• f(x) =
• 2x² + 13x + 20 =
• 2(x² + 13/2 x) + 20 =
• 2[x² + 2x*13/4 + (13/4)² - (13/4)²] + 20 =
• 2(x + 13/4)² - 169/8 + 20 =
• 2(x + 13/4)² - (169 - 8*20)/8 =
• 2(x + 13/4)² - 9/8

Answer 2

Answer:

$f(x)=2\left(x+\dfrac{13}{4}\right)^2-\dfrac{9}{8}$

Step-by-step explanation:

Given function:

$f(x)=2x^2+13x+20$

Factor out 2 from the terms in x:

$f(x)=2\left(x^2+\dfrac{13}{2}x\right)+20$

Add the square of half the coefficient of x inside the parentheses, and subtract the distributed equivalent outside the parentheses:

$f(x)=2\left(x^2+\dfrac{13}{2}x+\left(\dfrac{\frac{13}{2}}{2}\right)^2\right)+20-2\left(\dfrac{\frac{13}{2}}{2}\right)^2$

Simplify:

$f(x)=2\left(x^2+\dfrac{13}{2}x+\left(\dfrac{13}{4}\right)^2\right)+20-2\left(\dfrac{13}{4}\right)^2$

$f(x)=2\left(x^2+\dfrac{13}{2}x+\dfrac{169}{16}\right)+20-\dfrac{169}{8}$

$f(x)=2\left(x^2+\dfrac{13}{2}x+\dfrac{169}{16}\right)-\dfrac{9}{8}$

Factor the perfect trinomial inside the parentheses:

$f(x)=2\left(x+\dfrac{13}{4}\right)^2-\dfrac{9}{8}$