Question
what is the maximum of: y = 3x^2 - 12x + 18
Answers
Answered by
GPT-4o mini
The given function is a quadratic equation in the form \( y = ax^2 + bx + c \). Here, \( a = 3 \), \( b = -12 \), and \( c = 18 \).
Since the coefficient of \( x^2 \) (which is \( a \)) is positive, the parabola opens upwards, indicating that the function has a minimum value rather than a maximum value.
To find the vertex of the quadratic, which gives the minimum value, we can use the formula for the x-coordinate of the vertex:
\[
x = -\frac{b}{2a}
\]
Plugging in the values of \( a \) and \( b \):
\[
x = -\frac{-12}{2 \cdot 3} = \frac{12}{6} = 2
\]
Next, we'll substitute \( x = 2 \) back into the function to find the corresponding minimum value of \( y \):
\[
y = 3(2)^2 - 12(2) + 18
\]
Calculating this step by step:
\[
y = 3(4) - 24 + 18
\]
\[
y = 12 - 24 + 18
\]
\[
y = 6
\]
Thus, the minimum value of the function \( y = 3x^2 - 12x + 18 \) is 6, occurring at \( x = 2 \).
In summary:
- The function does not have a maximum value; it has a minimum value of 6 at \( x = 2 \).
Since the coefficient of \( x^2 \) (which is \( a \)) is positive, the parabola opens upwards, indicating that the function has a minimum value rather than a maximum value.
To find the vertex of the quadratic, which gives the minimum value, we can use the formula for the x-coordinate of the vertex:
\[
x = -\frac{b}{2a}
\]
Plugging in the values of \( a \) and \( b \):
\[
x = -\frac{-12}{2 \cdot 3} = \frac{12}{6} = 2
\]
Next, we'll substitute \( x = 2 \) back into the function to find the corresponding minimum value of \( y \):
\[
y = 3(2)^2 - 12(2) + 18
\]
Calculating this step by step:
\[
y = 3(4) - 24 + 18
\]
\[
y = 12 - 24 + 18
\]
\[
y = 6
\]
Thus, the minimum value of the function \( y = 3x^2 - 12x + 18 \) is 6, occurring at \( x = 2 \).
In summary:
- The function does not have a maximum value; it has a minimum value of 6 at \( x = 2 \).