To find the vertex of the quadratic function [tex]\( f(x) = 9x^2 + 1 \)[/tex], we follow these steps:
1. Identify the coefficients:
For a quadratic function in the form [tex]\( f(x) = ax^2 + bx + c \)[/tex], we have:
- [tex]\( a = 9 \)[/tex]
- [tex]\( b = 0 \)[/tex] (since there is no [tex]\(x\)[/tex] term)
- [tex]\( c = 1 \)[/tex]
2. Calculate the x-coordinate of the vertex:
The x-coordinate of the vertex, [tex]\( h \)[/tex], can be found using the formula:
[tex]\[
h = -\frac{b}{2a}
\][/tex]
Substituting the values of [tex]\( a \)[/tex] and [tex]\( b \)[/tex]:
[tex]\[
h = -\frac{0}{2 \times 9} = 0
\][/tex]
3. Calculate the y-coordinate of the vertex:
The y-coordinate of the vertex, [tex]\( k \)[/tex], is the value of the function evaluated at [tex]\( x = h \)[/tex]:
[tex]\[
k = f(h) = f(0)
\][/tex]
Substituting [tex]\( x = 0 \)[/tex] into the function:
[tex]\[
k = 9(0)^2 + 1 = 1
\][/tex]
4. Write the vertex coordinates:
Therefore, the vertex of the quadratic function [tex]\( f(x) = 9x^2 + 1 \)[/tex] is:
[tex]\[
(h, k) = (0, 1)
\][/tex]
In decimal form, the vertex is:
[tex]\[
(0.0, 1.0)
\][/tex]
So, the vertex of the function [tex]\( f(x) = 9x^2 + 1 \)[/tex] is [tex]\( (0.0, 1.0) \)[/tex].
