To find the vertex of the quadratic function [tex]\( y = 5x^2 + 20x + 1 \)[/tex], we can use the vertex form of a quadratic equation and some simple algebraic steps.
A quadratic equation of the form [tex]\( y = ax^2 + bx + c \)[/tex] has its vertex at the point [tex]\((x, y)\)[/tex], where:
[tex]\[ x = -\frac{b}{2a} \][/tex]
and
[tex]\[ y \][/tex] is the function value at that [tex]\( x \)[/tex].
1. First, let's identify the coefficients [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex] in the given function:
[tex]\[
a = 5, \quad b = 20, \quad c = 1
\][/tex]
2. Calculate the x-coordinate of the vertex using the formula [tex]\( x = -\frac{b}{2a} \)[/tex]:
[tex]\[
x = -\frac{20}{2 \cdot 5} = -\frac{20}{10} = -2
\][/tex]
3. To find the y-coordinate of the vertex, substitute [tex]\( x = -2 \)[/tex] back into the original quadratic function:
[tex]\[
y = 5(-2)^2 + 20(-2) + 1
\][/tex]
[tex]\[
y = 5 \cdot 4 + 20 \cdot (-2) + 1
\][/tex]
[tex]\[
y = 20 - 40 + 1
\][/tex]
[tex]\[
y = -19
\][/tex]
Therefore, the vertex of the quadratic function [tex]\( y = 5x^2 + 20x + 1 \)[/tex] is [tex]\((-2, -19)\)[/tex].
Among the given choices:
- [tex]\((-2, -19)\)[/tex]
- [tex]\((-19, -2)\)[/tex]
- [tex]\((19, 2)\)[/tex]
- [tex]\((2, 19)\)[/tex]
The correct answer is [tex]\((-2, -19)\)[/tex].
