Question

Determine the vertex of the function f(x) = 3x2 – 6x + 13. 1. Identify the values of a and b. a = and b = 2. Find the x-coordinate of the vertex. = 3. Find the y-coordinate by evaluating the function at the x-value found in the previous step. The vertex is.

Answer 1

F(x)=3x²-6x+13
a=3, b=-6, c=13
the x coordinate of the vertex is x=-b/(2a), so x=-(-6)/(2*3)=1
when x=1, y=3(1)²-6(1)+13=10
the vertex is at (1,10)

answers are in bold. 

Answer 2

Answer:

1) a = 3 , b = -6

2) x-coordinate of the vertex  is 1.

3) y-coordinate of the vertex is 10

Step-by-step explanation:

Given  function $f(x) = 3x^2-6x+ 13$

1)  Identify the values of a and b

The standard form of quadratic  function is  represented by $f(x)=ax^2+bx+c$

So on comparing with given quadratic function  $f(x) = 3x^2-6x+ 13$

We have a = 3  and b = - 6

2)  the x-coordinate of the vertex

Given function $f(x) = 3x^2-6x+ 13$

Writing in standard  form is $f(x)=(x-h)^2+k$ , where (h,k) is the vertex,

We first solve the given function $f(x) = 3x^2-6x+ 13$ by using completing square method, we have,

Taking 3 common , we get

$f(x) = 3(x^2-2x+\frac{13}{3})$

Using algebraic identity,  $(a-b)^2=a^2-2ab+b^2$

Here, a = 1 -2ab = -2 so b = 1

So add and subtract 1 both side, we get,

$f(x) = 3(x^2-2x+1-1+\frac{13}{3})$

Simplify we get,

$f(x) = 3((x-1)^2-1+\frac{13}{3})$

Solving , we get,

$f(x) = 3((x-1)^2+\frac{13-3}{3})$

$f(x) = 3((x-1)^2+\frac{10}{3})$

$f(x) =3((x-1)^2)+10$        

Thus, (h,k) = (1,10)

And thus x-coordinate of the vertex  is 1.

and y-coordinate of the vertex is 10