Answer :
To determine the coordinates of the vertex for the quadratic function [tex]\( f(x) = x^2 + 10x - 3 \)[/tex], we follow these steps:
1. Identify the coefficients:
The quadratic function is in the form [tex]\( ax^2 + bx + c \)[/tex], where:
- [tex]\( a = 1 \)[/tex]
- [tex]\( b = 10 \)[/tex]
- [tex]\( c = -3 \)[/tex]
2. Use the vertex formula:
The x-coordinate of the vertex of a quadratic function [tex]\( ax^2 + bx + c \)[/tex] is given by the formula:
[tex]\[ x = -\frac{b}{2a} \][/tex]
Substituting the values of [tex]\( a \)[/tex] and [tex]\( b \)[/tex]:
[tex]\[ x = -\frac{10}{2 \times 1} = -\frac{10}{2} = -5 \][/tex]
3. Find the y-coordinate by substituting [tex]\( x = -5 \)[/tex] back into the function:
[tex]\[ f(-5) = (-5)^2 + 10(-5) - 3 \][/tex]
[tex]\[ f(-5) = 25 - 50 - 3 \][/tex]
[tex]\[ f(-5) = -28 \][/tex]
So, the coordinates of the vertex are [tex]\((-5, -28)\)[/tex].
Therefore, the correct answer is:
[tex]\[ \boxed{(-5, -28)} \][/tex]
1. Identify the coefficients:
The quadratic function is in the form [tex]\( ax^2 + bx + c \)[/tex], where:
- [tex]\( a = 1 \)[/tex]
- [tex]\( b = 10 \)[/tex]
- [tex]\( c = -3 \)[/tex]
2. Use the vertex formula:
The x-coordinate of the vertex of a quadratic function [tex]\( ax^2 + bx + c \)[/tex] is given by the formula:
[tex]\[ x = -\frac{b}{2a} \][/tex]
Substituting the values of [tex]\( a \)[/tex] and [tex]\( b \)[/tex]:
[tex]\[ x = -\frac{10}{2 \times 1} = -\frac{10}{2} = -5 \][/tex]
3. Find the y-coordinate by substituting [tex]\( x = -5 \)[/tex] back into the function:
[tex]\[ f(-5) = (-5)^2 + 10(-5) - 3 \][/tex]
[tex]\[ f(-5) = 25 - 50 - 3 \][/tex]
[tex]\[ f(-5) = -28 \][/tex]
So, the coordinates of the vertex are [tex]\((-5, -28)\)[/tex].
Therefore, the correct answer is:
[tex]\[ \boxed{(-5, -28)} \][/tex]