To find the vertex of a quadratic function of the form [tex]\( y = ax^2 + bx + c \)[/tex], you can use the formula for the coordinates of the vertex. The vertex [tex]\((x_v, y_v)\)[/tex] is given by:
1. The x-coordinate of the vertex is found using the formula:
[tex]\[
x_v = -\frac{b}{2a}
\][/tex]
2. Once you have the x-coordinate, you substitute it back into the original quadratic equation to find the y-coordinate:
[tex]\[
y_v = a(x_v^2) + b(x_v) + c
\][/tex]
Given the quadratic equation [tex]\( y = x^2 + 6x + 5 \)[/tex]:
1. Identify the coefficients [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex]:
- [tex]\( a = 1 \)[/tex]
- [tex]\( b = 6 \)[/tex]
- [tex]\( c = 5 \)[/tex]
2. Calculate the x-coordinate of the vertex:
[tex]\[
x_v = -\frac{b}{2a} = -\frac{6}{2(1)} = -\frac{6}{2} = -3
\][/tex]
3. Substitute [tex]\( x_v = -3 \)[/tex] back into the quadratic equation to find [tex]\( y_v \)[/tex]:
[tex]\[
y_v = 1(-3)^2 + 6(-3) + 5
\][/tex]
[tex]\[
y_v = 1(9) + 6(-3) + 5
\][/tex]
[tex]\[
y_v = 9 - 18 + 5
\][/tex]
[tex]\[
y_v = 9 - 18 + 5 = -4
\][/tex]
Therefore, the vertex of the graph of the function [tex]\( y = x^2 + 6x + 5 \)[/tex] is [tex]\( (-3, -4) \)[/tex].
The correct answer is:
C. [tex]\((-3, -4)\)[/tex]
