Find the Maximum/Minimum Value -0.2t^2+1.6t+98.8

Solution:

Step 1:

The peak of a parabola $y=a{x}^2+bx+c$ 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 $y=a{x}^2+bx+c$. 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.