Answer :
To find the [tex]\( y \)[/tex]-value of the vertex of a quadratic function given by [tex]\( y = ax^2 + bx + c \)[/tex], follow these steps:
1. Identify the [tex]\( x \)[/tex]-value of the vertex:
The [tex]\( x \)[/tex]-value of the vertex of a quadratic function can be found using the formula:
[tex]\[ x = -\frac{b}{2a} \][/tex]
For typical values [tex]\( a = 1 \)[/tex], [tex]\( b = 1 \)[/tex], and [tex]\( c = 1 \)[/tex], this would give:
[tex]\[ x = -\frac{1}{2 \cdot 1} = -0.5 \][/tex]
2. Substitute the [tex]\( x \)[/tex]-value into the quadratic equation:
Once you have the [tex]\( x \)[/tex]-value of the vertex, substitute it back into the original quadratic equation to find the corresponding [tex]\( y \)[/tex]-value.
[tex]\[ y = a \left(-0.5 \right)^2 + b \left(-0.5 \right) + c \][/tex]
Using the values [tex]\( a = 1 \)[/tex], [tex]\( b = 1 \)[/tex], and [tex]\( c = 1 \)[/tex]:
[tex]\[ y = 1 \left(-0.5 \right)^2 + 1 \left(-0.5 \right) + 1 \][/tex]
Calculate each term individually:
[tex]\[ \left(-0.5 \right)^2 = 0.25 \][/tex]
[tex]\[ 1 \cdot 0.25 = 0.25 \][/tex]
[tex]\[ 1 \cdot -0.5 = -0.5 \][/tex]
[tex]\[ y = 0.25 - 0.5 + 1 \][/tex]
[tex]\[ y = 0.75 \][/tex]
3. Combine the results:
So, the [tex]\( y \)[/tex]-value of the vertex when [tex]\( a = 1 \)[/tex], [tex]\( b = 1 \)[/tex], and [tex]\( c = 1 \)[/tex] is:
[tex]\[ y = 0.75 \][/tex]
Thus, the coordinates of the vertex of the quadratic function given by [tex]\( y = ax^2 + bx + c \)[/tex] with values [tex]\( a = 1 \)[/tex], [tex]\( b = 1 \)[/tex], and [tex]\( c = 1 \)[/tex] are [tex]\( (-0.5, 0.75) \)[/tex].
1. Identify the [tex]\( x \)[/tex]-value of the vertex:
The [tex]\( x \)[/tex]-value of the vertex of a quadratic function can be found using the formula:
[tex]\[ x = -\frac{b}{2a} \][/tex]
For typical values [tex]\( a = 1 \)[/tex], [tex]\( b = 1 \)[/tex], and [tex]\( c = 1 \)[/tex], this would give:
[tex]\[ x = -\frac{1}{2 \cdot 1} = -0.5 \][/tex]
2. Substitute the [tex]\( x \)[/tex]-value into the quadratic equation:
Once you have the [tex]\( x \)[/tex]-value of the vertex, substitute it back into the original quadratic equation to find the corresponding [tex]\( y \)[/tex]-value.
[tex]\[ y = a \left(-0.5 \right)^2 + b \left(-0.5 \right) + c \][/tex]
Using the values [tex]\( a = 1 \)[/tex], [tex]\( b = 1 \)[/tex], and [tex]\( c = 1 \)[/tex]:
[tex]\[ y = 1 \left(-0.5 \right)^2 + 1 \left(-0.5 \right) + 1 \][/tex]
Calculate each term individually:
[tex]\[ \left(-0.5 \right)^2 = 0.25 \][/tex]
[tex]\[ 1 \cdot 0.25 = 0.25 \][/tex]
[tex]\[ 1 \cdot -0.5 = -0.5 \][/tex]
[tex]\[ y = 0.25 - 0.5 + 1 \][/tex]
[tex]\[ y = 0.75 \][/tex]
3. Combine the results:
So, the [tex]\( y \)[/tex]-value of the vertex when [tex]\( a = 1 \)[/tex], [tex]\( b = 1 \)[/tex], and [tex]\( c = 1 \)[/tex] is:
[tex]\[ y = 0.75 \][/tex]
Thus, the coordinates of the vertex of the quadratic function given by [tex]\( y = ax^2 + bx + c \)[/tex] with values [tex]\( a = 1 \)[/tex], [tex]\( b = 1 \)[/tex], and [tex]\( c = 1 \)[/tex] are [tex]\( (-0.5, 0.75) \)[/tex].
