Respuesta :
Answer:
The vertex form of a quadratic equation is:
[tex]f(x) = (x-3)^2+4[/tex]
the minimum value of f(x) is [tex]y=4[/tex]
Step-by-step explanation:
Given a quadratic equation of the form [tex]f (x) = ax ^ 2 + bx + c[/tex] then the x coordinate of the vertex is
[tex]x=-\frac{b}{2a}[/tex]
So for [tex]f(x)=x^2-6x+13[/tex]
[tex]a=1\\b=-6\\c=13\\[/tex]
Therefore
The x coordinate of the vertex is:
[tex]x=-\frac{(-6)}{2(1)}[/tex]
[tex]x=3[/tex]
The y coordinate of the vertex is:
[tex]f(3)=(3)^2-6(3)+13[/tex]
[tex]y=f(3)=4[/tex]
By definition the minimum value of the quadratic function is the same as the coordinate of y of its vertex
So the minimum value is [tex]y=4[/tex]
The vertex form of a quadratic equation is:
[tex]f(x) = a(x-h)^2+k[/tex]
Where
a is the main coefficient. [tex]a=1[/tex]
h is the x coordinate of the vertex. [tex]h=3[/tex]
k is the y coordinate of the vertex. [tex]k=4[/tex]
So the vertex form of a quadratic equation is:
[tex]f(x) = (x-3)^2+4[/tex]
Answer:
a. [tex]f(x)=(x-3)^2+4[/tex]
b. The minimum value is 4
Step-by-step explanation:
The given function is: [tex]f(x)=x^2-6x+13[/tex]
We add and subtract half the square of the coefficient of x.
[tex]f(x)=x^2-6x+3^2-3^2+13[/tex]
This becomes: [tex]f(x)=x^2-6x+9-9+13[/tex]
The first three terms form a perfect square trinomial.
[tex]f(x)=(x-3)^2+4[/tex]
The function is now in the form: [tex]f(x)=a(x-h)^2+k[/tex], where V(h,k) is the vertex.
Therefore the vertex is (3,4).
The minimum value is the y-value of the vertex, which is 4.