Question
Find the minimum value of the parabola y = 48/5x^2.
Simplify your answer and write it as a proper fraction, improper fraction, or integer.
Simplify your answer and write it as a proper fraction, improper fraction, or integer.
Answers
GPT-4o mini
The given function is \( y = \frac{48}{5} x^2 \).
This function is a quadratic in the form \( y = ax^2 + bx + c \), where \( a = \frac{48}{5} \), \( b = 0 \), and \( c = 0 \).
Since \( a > 0 \), this is a parabola that opens upwards. The minimum value occurs at the vertex of the parabola.
For a quadratic function in the form \( y = ax^2 + bx + c \), the \( x \)-coordinate of the vertex is given by:
\[
x = -\frac{b}{2a}
\]
Substituting \( b = 0 \) and \( a = \frac{48}{5} \):
\[
x = -\frac{0}{2 \cdot \frac{48}{5}} = 0
\]
Now, we evaluate \( y \) at \( x = 0 \):
\[
y = \frac{48}{5} (0)^2 = 0
\]
Thus, the minimum value of the parabola is \( 0 \).
The minimum value of the parabola \( y = \frac{48}{5} x^2 \) is
\[
\boxed{0}
\]
This function is a quadratic in the form \( y = ax^2 + bx + c \), where \( a = \frac{48}{5} \), \( b = 0 \), and \( c = 0 \).
Since \( a > 0 \), this is a parabola that opens upwards. The minimum value occurs at the vertex of the parabola.
For a quadratic function in the form \( y = ax^2 + bx + c \), the \( x \)-coordinate of the vertex is given by:
\[
x = -\frac{b}{2a}
\]
Substituting \( b = 0 \) and \( a = \frac{48}{5} \):
\[
x = -\frac{0}{2 \cdot \frac{48}{5}} = 0
\]
Now, we evaluate \( y \) at \( x = 0 \):
\[
y = \frac{48}{5} (0)^2 = 0
\]
Thus, the minimum value of the parabola is \( 0 \).
The minimum value of the parabola \( y = \frac{48}{5} x^2 \) is
\[
\boxed{0}
\]