How do you write $f (x) = 3 x^{2} + 6 x + 1$ $f(x) = 3x^2 + 6x + 1$ in vertex form?

Question

How do you write $f (x) = 3 x^{2} + 6 x + 1$ $f(x) = 3x^2 + 6x + 1$ in vertex form?

1 Answer

Impact of this question

3023 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-08-24T04:38:16

$f (x) = 3 {(x + 1)}^{2} - 2$ $f(x) = 3(x + 1)^2 - 2$

Explanation:

Complete the square:

$f (x) = 3 (x^{2} + 2 x + m - m) + 1$ $f(x) = 3(x^2 + 2x + m - m) + 1$

To find $m$ $m$ we use $m = {(\frac{b}{2})}^{2} = {(\frac{2}{2})}^{2} = 1$ $m = (b/2)^2 = (2/2)^2 = 1$

$f (x) = 3 (x^{2} + 2 x + 1 - 1) + 1$ $f(x) = 3(x^2 + 2x + 1- 1) + 1$

$f (x) = 3 (x^{2} + 2 x + 1) - 3 + 1$ $f(x) = 3(x^2 + 2x + 1) - 3 + 1$

$f (x) = 3 {(x + 1)}^{2} - 2$ $f(x) = 3(x + 1)^2 - 2$

Hopefully this helps!

Science

Math

Humanities

... and beyond

How do you write $f (x) = 3 x^{2} + 6 x + 1$ $f(x) = 3x^2 + 6x + 1$ in vertex form?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you write f(x)=3x2+6x+1f(x) = 3x^2 + 6x + 1 in vertex form?

1 Answer

Explanation:

Related questions

Impact of this question

How do you write $f (x) = 3 x^{2} + 6 x + 1$ $f(x) = 3x^2 + 6x + 1$ in vertex form?