91Ó°ÊÓ

The axis of symmetry of a quadratic function is a vertical line that divides the parabola into two mirror images. It's found right in the middle of the two real zeros of the function, if they exist. For the function \( P(s) = -2s^2 - 4s + 48 \), we use the formula: \( s = -\frac{b}{2a} \). Here, \( a \) and \( b \) are coefficients from the standard quadratic form \( as^2 + bs + c \). \( a \) is -2 and \( b \) is -4, so substituting in these values, we get: \( s = -\frac{-4}{2(-2)} = \frac{-4}{-4} = 1 \). So, the axis of symmetry for this quadratic function is at \( s = 1 \). This means any vertical line drawn at \( s = 1 \) will perfectly divide the parabola into two symmetric parts.

The vertex of a parabola is either its highest or lowest point, depending on whether it opens downward or upward. For the quadratic function \( P(s) = -2s^2 - 4s + 48 \), it opens downward because the leading coefficient \( a = -2 \) is negative.
To find the vertex, you start with the axis of symmetry. In this case, the axis of symmetry is at \( s = 1 \). Substitute \( s = 1 \) into \( P(s) \) to find the y-coordinate of the vertex: \( P(1) = -2(1)^2 - 4(1) + 48 \)\( P(1) = -2 - 4 + 48 = 42 \).
The coordinates of the vertex are \( (1, 42) \). This means the highest point on the graph of the quadratic function is at 1 on the \( s \)-axis and 42 on the \( P(s) \)-axis.

Real zeros are the points where the graph of the quadratic function intersects the horizontal \( s \)-axis. For a quadratic function \( P(s) = a(s - r_1)(s - r_2) \), the real zeros are \( r_1 \) and \( r_2 \). In our example, the given zeros are \( s = -6 \) and \( s = 4 \).
To confirm these real zeros, let's consider the constructed function: \( P(s) = -2(s + 6)(s - 4) \). Setting this equation to zero gives us:

• \( P(-6) = -2((-6) + 6)(-6 - 4) = 0 \)
• \( P(4) = -2(4 + 6)(4 - 4) = 0 \)

Both calculations demonstrate that the real zeros are exactly \( -6 \) and \( 4 \), where the graph touches the \( s \)-axis.

The vertical intercept, also known as the y-intercept, is where the graph of the function crosses the vertical \( P(s) \)-axis. This happens when \( s = 0 \). To find it, substitute \( s = 0 \) into the quadratic function. For \( P(s) = -2s^2 - 4s + 48 \), this is:\( P(0) = -2(0)^2 - 4(0) + 48 \)\( P(0) = 48 \).
The vertical intercept is \( 48 \). This means when \( s = 0 \), \( P(s) = 48 \), so the function crosses the vertical axis at \( 48 \). The graph therefore intersects the \( P(s) \)-axis at the point \( (0, 48) \).

A quadratic equation is a polynomial equation of degree two, which can be generally expressed as \( as^2 + bs + c = 0 \). This equation represents a parabola when graphed.
To construct a specific quadratic function like we did, we use the form \( P(s) = a(s - r_1)(s - r_2) \) where \( r_1 \) and \( r_2 \) are the real zeros of the function. In this case, we had zeros at \( s = -6 \) and \( s = 4 \), which gave us: \( P(s) = a(s + 6)(s - 4) \).
Given a point \( (5, -22) \) on the function, we determined \( a = -2 \). Expanding this, we got:\( P(s) = -2(s^2 + 2s - 24) = -2s^2 - 4s + 48 \). This is our specific quadratic function and shows how the parabola behaves in relation to its square, linear, and constant terms.