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PLEASE HELP Let f (x) = x2 – 6x + 13.What is the vertex form off (x)?What is the minimum value off (x)?Enter your answers in the boxes.Vertex form: f (x) = 1Minimum value of f (x):

PLEASE HELP Let f (x) = x2 – 6x + 13.What is the vertex form off (x)?What is the minimum-example-1

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The function f(x) is a quadratic function and we can define, for a quadratic function, its vertex form as follows:


y=a(x-h)^2+k

Where a, h, and k are numbers

Notice that if we expand the previous expression, we get:


a(x-h)^2+k=ax^2-ahx+ah^2+k

So, from the function f(x)=x^2-6x+13 we can see that a=1 and that a*h=6, i.e.,


ah=6,a=1\Rightarrow h=6

Furthermore, notice that:


ah^2+k=13\Rightarrow k=13-ah^2

and h=6, a=1!


\Rightarrow k=13-36=-23

We have everything we need to write the vertex form:


f(x)=1\cdot(x-6)^2-23

Now, we can calculate its minimum value!

Looking at the function we can notice that its graph is a parabola that opens towards the +y direction (it has a 'U' shape). Therefore, there is a point that is the lowest point of the graph!

We can calculate it by means of the derivative:


f^(\prime)(x)=2x-6

And, to find the minimum we do f'(x)=0


f^(\prime)(x)=0\Rightarrow2x-6=0\Rightarrow x=3

And x=3 gives us the minimum value of f(x)! In this way:


f(3)=3^2-6(3)+13=9-18+13=4

User Kubrick G
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