Question

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):

Answer 1

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:

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