Find the Maximum/Minimum Value -0.2t^2+1.6t+98.8
The problem presents a quadratic equation in the form of -0.2t^2 + 1.6t + 98.8 and asks for the maximum or minimum value of this quadratic function. This task involves determining the vertex of the parabola represented by the equation, which is where the maximum or minimum value (depending on the direction the parabola opens) is located. Since the coefficient of the t^2 term is negative, the parabola opens downwards, which means the vertex will represent the maximum value of the function. The question seeks that maximum value without actually requiring the calculation or derivation process.
$- 0.2 t^{2} + 1.6 t + 98.8$
Answer
Solution:
Step 1:
The peak of a parabola $y = ax^2 + bx + c$ is found at $x = -\frac{b}{2a}$. When $a < 0$, the parabola opens downward and the peak represents the maximum value, which is $y_{\text{max}}$ at $x = -\frac{b}{2a}$.
Step 2:
Determine the $x$-coordinate of the vertex, $x = -\frac{b}{2a}$.
Step 2.1:
Insert the known coefficients $a$ and $b$ into the formula: $x = -\frac{1.6}{2(-0.2)}$.
Step 2.2:
Eliminate the brackets: $x = -\frac{1.6}{2(-0.2)}$.
Step 2.3:
Perform the calculation of $-\frac{1.6}{2(-0.2)}$.
Step 2.3.1:
Calculate $2 \times -0.2$: $x = -\frac{1.6}{-0.4}$.
Step 2.3.2:
Divide $1.6$ by $-0.4$: $x = 4$.
Step 2.3.3:
The result is $x = 4$.
Step 3:
Compute $f(4)$.
Step 3.1:
Substitute $x$ with $4$ in the quadratic equation: $f(4) = -0.2(4)^2 + 1.6(4) + 98.8$.
Step 3.2:
Simplify the expression.
Step 3.2.1:
Break down the expression term by term.
Step 3.2.1.1:
Square the number $4$: $f(4) = -0.2 \cdot 16 + 1.6(4) + 98.8$.
Step 3.2.1.2:
Multiply $-0.2$ by $16$: $f(4) = -3.2 + 1.6(4) + 98.8$.
Step 3.2.1.3:
Multiply $1.6$ by $4$: $f(4) = -3.2 + 6.4 + 98.8$.
Step 3.2.2:
Combine the numerical values.
Step 3.2.2.1:
Add $-3.2$ to $6.4$: $f(4) = 3.2 + 98.8$.
Step 3.2.2.2:
Add $3.2$ to $98.8$: $f(4) = 102$.
Step 3.2.3:
The maximum value of the function is $102$.
Step 4:
The maximum point on the graph is at $(4, 102)$.
Knowledge Notes:
The problem involves finding the maximum value of a quadratic function, which is a polynomial of degree two, typically written in the form $y = ax^2 + bx + c$. The key features of a quadratic function include its vertex, axis of symmetry, and whether it opens upwards or downwards, which is determined by the sign of the coefficient $a$. If $a$ is positive, the parabola opens upwards, and the vertex is the minimum point. If $a$ is negative, the parabola opens downwards, and the vertex is the maximum point.
The vertex of a parabola can be found using the vertex formula $x = -\frac{b}{2a}$. Once the $x$-coordinate of the vertex is known, it can be substituted back into the original equation to find the corresponding $y$-coordinate, which gives the maximum or minimum value of the function.
In this specific problem, the coefficient $a$ is negative, indicating that the parabola opens downwards and thus has a maximum value. By following the steps outlined in the solution, we find the maximum value and the point at which it occurs on the graph of the function.
