The given function is \( h(t) = 2t^2 + 3t - 2 \). This represents a quadratic function, which typically forms a parabola when graphed. Let's analyze this function step by step.

### 1. General Form
This quadratic function is in the standard form \( h(t) = at^2 + bt + c \), where:
- \( a = 2 \)
- \( b = 3 \)
- \( c = -2 \)

### 2. Vertex
The vertex of a parabola can be found using the formula for the \( t \)-coordinate of the vertex:
\[
t_{\text{vertex}} = -\frac{b}{2a}
\]
Substitute the values of \( a \) and \( b \):
\[
t_{\text{vertex}} = -\frac{3}{2(2)} = -\frac{3}{4}
\]
Now, plug this value back into the function to find the corresponding \( h(t) \)-value:
\[
h\left(-\frac{3}{4}\right) = 2\left(-\frac{3}{4}\right)^2 + 3\left(-\frac{3}{4}\right) - 2
\]
\[
= 2\left(\frac{9}{16}\right) - \frac{9}{4} - 2
\]
\[
= \frac{18}{16} - \frac{9}{4} - 2 = \frac{9}{8} - \frac{9}{4} - 2
\]
Simplifying:
\[
\frac{9}{8} - \frac{18}{8} = -\frac{9}{8}, \quad \text{so the full expression is} \quad -\frac{9}{8} - 2 = -\frac{9}{8} - \frac{16}{8} = -\frac{25}{8}
\]
Thus, the vertex is \( \left( -\frac{3}{4}, -\frac{25}{8} \right) \).

### 3. Axis of Symmetry
The axis of symmetry is the vertical line that passes through the vertex, so it is:
\[
t = -\frac{3}{4}
\]

### 4. Direction of the Parabola
Since \( a = 2 \) is positive, the parabola opens upwards.

### 5. \( h(t) \)-intercept (Y-intercept)
The \( h(t) \)-intercept occurs when \( t = 0 \). Substituting \( t = 0 \) into the function:
\[
h(0) = 2(0)^2 + 3(0) - 2 = -2
\]
Thus, the \( h(t) \)-intercept is \( (0, -2) \).

### 6. Roots (t-intercepts)
To find the roots, set \( h(t) = 0 \):
\[
2t^2 + 3t - 2 = 0
\]
Solve this quadratic equation using the quadratic formula:
\[
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Substitute the values of \( a = 2 \), \( b = 3 \), and \( c = -2 \):
\[
t = \frac{-3 \pm \sqrt{3^2 - 4(2)(-2)}}{2(2)} = \frac{-3 \pm \sqrt{9 + 16}}{4} = \frac{-3 \pm \sqrt{25}}{4}
\]
\[
t = \frac{-3 \pm 5}{4}
\]
Thus, the two solutions are:
\[
t_1 = \frac{-3 + 5}{4} = \frac{2}{4} = \frac{1}{2}, \quad t_2 = \frac{-3 - 5}{4} = \frac{-8}{4} = -2
\]
Therefore, the roots are \( t = \frac{1}{2} \) and \( t = -2 \).

Would you like further details or have any questions?

Here are 5 related questions:
1. How does the vertex of a quadratic function relate to its maximum or minimum value?
2. Can you explain how the quadratic formula works for finding roots?
3. What would happen if the parabola opened downward instead of upward?
4. How can you graph this quadratic function?
5. What are the different forms of quadratic functions, and how do they help in solving problems?

**Tip:** When solving quadratics, always check the discriminant (\( b^2 - 4ac \)) first to determine the number of real roots.

Explore how to solve the quadratic equation h(t) = 2t² + 3t - 2. This comprehensive solution covers key algebraic concepts such as finding the vertex, axis of symmetry, and roots using the quadratic formula. Ideal for students in Grades 9-10.

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