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In quadratic functions like the one given, the vertex is a key point that tells us a lot about the graph of the function. For our function expressed as \( s(t) = -9t^2 + 3t + 2 \), the vertex helps us understand where the function reaches its maximum value because our parabola opens downward. We find the vertex using the formula \( t = \frac{-b}{2a} \). Substituting the coefficients \( b = 3 \) and \( a = -9 \), we get \( t = \frac{1}{6} \).

To find the corresponding \( s \)-value, we substitute \( t = \frac{1}{6} \) back into the function and calculate \( s = \frac{25}{12} \). Therefore, the vertex can be located at the point \( \left(\frac{1}{6}, \frac{25}{12}\right) \). This point represents the highest point on the graph since the parabola opens downwards.

The axis of symmetry is a crucial feature of a quadratic graph and provides a mirror line that divides the parabola into two matching halves. For our function \( s(t) = -9t^2 + 3t + 2 \), like any standard quadratic equation \( ax^2 + bx + c \), the axis of symmetry can be found with the same formula used for the vertex: \( t = \frac{-b}{2a} \).

By calculating this, we determined \( t = \frac{1}{6} \). This equation is that of a vertical line passing through the vertex on a graph. The axis of symmetry helps in understanding that each side of this vertical line reflects the other side, making it symmetrical.

In a quadratic function where the parabola opens downward, the vertex represents the maximum value of the function. Our parabola \( s(t) = -9t^2 + 3t + 2 \) opens downward because the coefficient \( a = -9 \) is negative. Hence, the vertex \( \left(\frac{1}{6}, \frac{25}{12}\right) \) indicates not only a point but also the maximum value of the quadratic function.

So, the maximum value of the function is \( s = \frac{25}{12} \). This means that when \( t = \frac{1}{6} \), \( s \) reaches its highest point of \( \frac{25}{12} \) before the function value begins to decrease as you move towards either side along the \( t \)-axis.

Intercepts are points where the graph crosses the axes, providing significant insight into the behavior of the quadratic function. For the y-intercept, we set \( t = 0 \) in our function, resulting in \( s = 2 \). Therefore, the y-intercept of this quadratic is \( (0, 2) \). This is where the graph crosses the y-axis.

t-intercepts are often called roots or zeros and they are found by solving \( -9t^2 + 3t + 2 = 0 \). Using the quadratic formula, \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), we find that \( t \) can be \( \frac{2}{3} \) or \( -\frac{1}{3} \). This implies the graph intersects the t-axis at \( \left(-\frac{1}{3}, 0\right) \) and \( \left(\frac{2}{3}, 0\right) \). These intercepts indicate where the function value is zero, and they are crucial for sketching the graph of the function.