# VertexWhere is located the vertex of the graph of f(x)=x^2-8x+16 ?

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### 2 Answers

You need to determine where the vertex of quadratic function is located, with respect to x and y axis.

You also may determine the vertex using derivative of the function, such that:

`f'(x) = (x^2-8x+16)' => f'(x) = 2x - 8`

The x coordinate of vertex is the solution to equation `f'(x) = 0` , such that:

`2x - 8 = 0 => 2x = 8 => x = 8/2 => x = 4`

You may determine y coordinate of vertex, replacing 4 for x in quadratic equation of `f(x)` , such that:

`f(4) = 4^2-8*4+16 => f(4) = 16 - 32 + 16 => f(4) = 0`

**You should notice that the graph of the function `f(x) = x^2-8x+16` is tangent to x axis at `x = 4` .**

To find out the place the vertex of the parabola f(x) = y is, we'll have to identify the quadrant or the side of x axis where the coordinates of the vertex of the graph of parabola lie.

We'll recall that the coordinates of the parabola vertex are:

V(-b/2a;-delta/4a), where a,b,c are the coefficients of the function and delta=b^2 -4*a*c.

y=f(x)=x^2 - 8x + 16

We'll identify the coefficients:

a=1, 2a=2, 4a=4

b=-8, c=16

delta=(-8)^2 -4*1*16

delta =64 - 64

delta = 0

V(-b/2a;-delta/4a)=V(-(-8)/2;-(0)/4)

V(4;0)

Since the x coordinate is positive and y coordinate is 0, the vertex is located on the positive side of x axis: V(4;0).