The quadratic function given is:

$h(t) = -4t^2 + 2t - 1$

Maximum or Minimum:

• Since the coefficient of $t^2$ is negative (−4), the parabola opens downward, meaning the function has a maximum value.

Maximum Value:

The maximum value occurs at the vertex of the parabola. The formula for the vertex of a quadratic function $at^2 + bt + c$ is:

$t = -\frac{b}{2a}$

For $h(t) = -4t^2 + 2t - 1$, $a = -4$ and $b = 2$. Plugging these values into the vertex formula:

$t = -\frac{2}{2(-4)} = \frac{2}{8} = \frac{1}{4}$

So, the axis of symmetry is $t = \frac{1}{4}$.

Finding the Maximum Value of $h$:

Now, substitute $t = \frac{1}{4}$ back into the function to find the maximum value:

$h\left(\frac{1}{4}\right) = -4\left(\frac{1}{4}\right)^2 + 2\left(\frac{1}{4}\right) - 1$ $h\left(\frac{1}{4}\right) = -4\left(\frac{1}{16}\right) + \frac{1}{2} - 1$ $h\left(\frac{1}{4}\right) = -\frac{4}{16} + \frac{1}{2} - 1$ $h\left(\frac{1}{4}\right) = -\frac{1}{4} + \frac{1}{2} - 1 = \frac{-1 + 2 - 4}{4} = \frac{-3}{4}$

Thus, the maximum value of $h$ is −3/4, which you correctly found.

Axis of Symmetry:

The axis of symmetry is $t = \frac{1}{4}$, which should be the correct answer.

Would you like any additional details or clarification on this?

Here are five related questions to consider:

1. How would you find the vertex of a quadratic function that opens upward?
2. What is the significance of the discriminant in determining the number of real roots of a quadratic function?
3. How does the coefficient of $t^2$ affect the direction of the parabola?
4. How can we solve for the x-intercepts of this quadratic function?
5. What are some real-world applications of finding the maximum or minimum of a quadratic function?

Tip: Always check the sign of the quadratic term ($a$) to determine if the parabola opens upwards (minimum) or downwards (maximum).