# How do you find the vertex of y=x^2+4x+1?

Apr 20, 2018

The vertex is $\left(- 2 , - 3\right)$.

#### Explanation:

Note: when the variables a, b, c, etc. are used, I am referring to a general rule that will work for every real value of a, b, c, etc.

The vertex can be found in many ways:

The simplest is using a graphing calculator and finding the vertex that way- but I assume you mean how to calculate it mathematically:

In an equation $y = a {x}^{2} + b x + c$, the x value of the vertex is (-b)/(2a. (This can be proven, but I won't do that here to save some time).

Using the equation $y = {x}^{2} + 4 x + 1$, you can see that $a = 1 , b = 4 ,$ and $c = 1$. Therefore, the x value of the vertex is -4/(2(1), or $- 2$.

You can then plug that into the equation and solve for the y value of the vertex:

$y = {\left(- 2\right)}^{2} + 4 \left(- 2\right) + 1$; $y = 4 - 8 + 1$; $y = - 3$.

Therefore, the answer is $\left(- 2 , - 3\right)$.

Alternatively, you can solve by completing the square:

with $y = a {x}^{2} + b x + c$, you try to turn the equation into $y = {\left(x - d\right)}^{2} + f$, where the vertex is $\left(d , f\right)$. This is vertex form.

You have $y = {x}^{2} + 4 x + 1$. To complete the square, add 4 to both sides:

$y + 4 = {x}^{2} + 4 x + 4 + 1$.

I did this because ${x}^{2} + 4 x + 4$ is equal to ${\left(x + 2\right)}^{2}$, which is what we want to convert this into vertex form:

$y + 4 = {\left(x + 2\right)}^{2} + 1$

You can then subtract 4 from both sides to isolate $y$:

y=(x+2)^2+1-4; y=(x+2)^2-3.

With the form $y = {\left(x - d\right)}^{2} + f$ and vertex $\left(d , f\right)$, you can then see that the vertex is #(-2,-3).

graph{y=x^2+4x+1 [-10, 10, -5, 5]}

Hope this helps!