# Vertex.Write the vertex of the parabola y=x^2+4x.

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### 2 Answers

You need to know that the vertex of parabola represents an extreme point of the function `y = x^2 + 4x,` hence, you may use derivative of the function to find the extreme point, such that:

`(dy)/(dx) = (d(x^2 + 4x))/(dx) => (dy)/(dx) = 2x + 4`

You need to solve for x the equation `(dy)/(dx) = 0` to find the critical value, such that:

`2x + 4 = 0 => 2x = -4 => x = -4/2 => x = -2`

Since x = -2 represents the critical value, you may evaluate the extreme point of the function, such that:

`f(-2) = (-2)^2 + 4*(-2) => f(-2) = 4 - 8 => f(-2) = -4`

Hence, evaluating the vertex of the parabola `y = x^2 + 4x` yields `(h,k) = (-2,-4).`

We'll write the coordinates of the vertex of the parabola:

V(-b/2a ; -delta/4a)

a,b,c are the coefficients of the quadratic:

y = ax^2 + bx + c

Comparing, we'll identify the coefficients:

a = 1, b = 4 , c = 0

We'll calculate the x coordinate of the vertex:

xV = -b/2a

xV = -4/2

xV = -2

We'll calculate the y coordinate of the vertex:

yV = -delta/4a

delta = b^2 - 4ac

delta = 16 - 4*1*0

delta = 16

yV = -16/4

yV = -4

**The coordinates of the vertex of the parabola are: V(-2 , -4).**