Question

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

Answer 1

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

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=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!