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

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Answer 1 · 2018-04-20T03:26:35

The vertex is $(-2,-3)$ .

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 $y=ax^2+bx+c$ , 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 $y=x^2+4x+1$ , 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:

$y=(-2)^2+4(-2)+1$ ; $y=4-8+1$ ; $y=-3$ .

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

Alternatively, you can solve by completing the square:

with $y=ax^2+bx+c$ , you try to turn the equation into $y=(x-d)^2+f$ , where the vertex is $(d,f)$ . This is vertex form.

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

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

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

$y+4=(x+2)^2+1$

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 $y=(x-d)^2+f$ and vertex $(d,f)$ , you can then see that the vertex is #(-2,-3).

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

Hope this helps!

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