SOLUTION: find the cordinate of the vertex of the graph of f(x) f(x)=x2-4x+ 13

Question 1199166: find the cordinate of the vertex of the graph of f(x) f(x)=x2-4x+ 13
Found 2 solutions by math_tutor2020, josgarithmetic:
Answer by math_tutor2020(3817) (Show Source): You can put this solution on YOUR website!

f(x) = x^2-4x+13
is the given equation

Think of it as
y = x^2-4x+13

Compare it to
y = ax^2+bx+c

We have the following coefficients
a = 1
b = -4
c = 13

The vertex is located at (h,k).
Let's compute the x coordinate of the vertex.
h = -b/(2a)
h = -(-4)/(2*1)
h = 2

Then plug this into the original equation to find the y coordinate of the vertex.
y = x^2-4x+13
y = 2^2-4*2+13
y = 4-8+13
y = -4+13
y = 9
This is the value of k in (h,k)

Since
h = 2
k = 9
We can then say the vertex is located at (h,k) = (2,9)
This is the lowest point on the parabola since a = 2 is positive, causing the parabola to open upward.

The vertex form would be:
y = a(x-h)^2+k
y = 1(x-2)^2+9
y = (x-2)^2+9
I'll let the student expand that vertex form out so you get the original equation again.
This helps confirm the correct vertex form.

--------------------------------------------

Answer: The vertex is located at (2,9)

Answer by josgarithmetic(39620) (Show Source): You can put this solution on YOUR website!
Complete The Square for

------------------Vertex is at (2,9) which is read directly from the expression (for y or f(x)).
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