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How do you write this equation into the vertex form #f(x) = x^2+4x+6#?

Answer 1

Elias Ackerman

#f(x)=(x+2)^2+2#

#"the equation of a parabola in "color(blue)"vertex form"# is.

#color(red)(bar(ul(|color(white)(2/2)color(black)(y=a(x-h)^2+k)color(white)(2/2)|)))#

#"where "(h,k)" are the coordinates of the vertex and a"# #"is a multiplier"#

#"to obtain this form "color(blue)"complete the square"#

#f(x)=x^2+4x color(red)(+4)color(red)(-4)+6#

#color(white)(f(x))=(x+2)^2+2larrcolor(blue)"in vertex form"#

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Answer 2

Ariana Alvarez

To write the equation (f(x) = x^2 + 4x + 6) into vertex form, you complete the square.

First, factor out the coefficient of (x^2) (which is 1):

(f(x) = 1(x^2 + 4x) + 6)

Then, add and subtract ((4/2)^2 = 4) inside the parentheses:

(f(x) = 1(x^2 + 4x + 4 - 4) + 6)

Next, group the perfect square trinomial and the constant term:

(f(x) = 1[(x^2 + 4x + 4) - 4] + 6)

Now, rewrite the perfect square trinomial as a squared binomial:

(f(x) = 1[(x + 2)^2 - 4] + 6)

Finally, distribute the 1, simplify, and rewrite the equation in vertex form:

(f(x) = (x + 2)^2 - 4 + 6)

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

So, the equation in vertex form is (f(x) = (x + 2)^2 + 2).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.