Nathanial Frost

2023-02-18

How to write $f\left(x\right)={x}^{2}+4x+1$ in vertex form?

Rhys Murphy

Beginner2023-02-19Added 3 answers

The equation in vertex form is $f\left(x\right)=a{(x-h)}^{2}+k$

Here, we have $f\left(x\right)={x}^{2}+4x+1$, so we have $a=1$

So for converting to this form, we complete the square using $(x+a)}^{2}={x}^{2}+2ax+{a}^{2$ for which we have to identify $a$ and then add and subtract $a}^{2$ Thus,

$f\left(x\right)=\underline{{x}^{2}+2\times 2\times x+{2}^{2}}-{2}^{2}+1$

$={(x+2)}^{2}-4+1$

$={(x-(-2))}^{2}-3$

Here, we have $f\left(x\right)={x}^{2}+4x+1$, so we have $a=1$

So for converting to this form, we complete the square using $(x+a)}^{2}={x}^{2}+2ax+{a}^{2$ for which we have to identify $a$ and then add and subtract $a}^{2$ Thus,

$f\left(x\right)=\underline{{x}^{2}+2\times 2\times x+{2}^{2}}-{2}^{2}+1$

$={(x+2)}^{2}-4+1$

$={(x-(-2))}^{2}-3$