Question

15 find the maximum or minimum value of the function f(x) = -x² + 24x - 146

find the maximum or minimum value of the function f(x) = (x + 3)² + 6

20 find the quadratic function satisfying the following conditions: the axis of symmetry is x = 8, the y - intercept is (0, 64), there is only one x - intercept

find the quadratic function satisfying: the graph passes through the origin, the vertex is (10, 2)

Explanation:

Problem 19: Find the maximum or minimum value of \( f(x) = -x^2 + 24x - 146 \)

Step 1: Identify the vertex x-coordinate

For a quadratic function \( f(x) = ax^2 + bx + c \), the x-coordinate of the vertex is \( x = \frac{-b}{2a} \). Here, \( a = -1 \), \( b = 24 \).
\( x = \frac{-24}{2(-1)} = \frac{-24}{-2} = 12 \)

Step 2: Find the y-coordinate (function value at vertex)

Substitute \( x = 12 \) into \( f(x) \):
\( f(12) = -(12)^2 + 24(12) - 146 \)
\( = -144 + 288 - 146 \)
\( = ( -144 - 146 ) + 288 \)
\( = -290 + 288 = -2 \)

Since \( a = -1 < 0 \), the parabola opens downward, so the vertex is a maximum point.

The function is in vertex form \( f(x) = a(x - h)^2 + k \), where \( (h, k) \) is the vertex. Here, \( a = 1 \), \( h = -3 \), \( k = 6 \).

Since \( a = 1 > 0 \), the parabola opens upward, so the vertex is a minimum point. The minimum value is \( k = 6 \) (occurring at \( x = -3 \)).

Step 1: Vertex form with axis of symmetry

The axis of symmetry is \( x = 8 \), so the vertex has x-coordinate \( h = 8 \). Since there's one x-intercept, the vertex lies on the x-axis, so the vertex is \( (8, 0) \). Thus, the function is \( f(x) = a(x - 8)^2 + 0 = a(x - 8)^2 \).

Step 2: Use y-intercept to find \( a \)

The y-intercept is \( (0, 64) \), so substitute \( x = 0 \), \( f(x) = 64 \):
\( 64 = a(0 - 8)^2 \)
\( 64 = a(64) \)
\( a = 1 \)

Thus, the function is \( f(x) = (x - 8)^2 = x^2 - 16x + 64 \).

Answer:

The maximum value of the function is \( -2 \) (occurring at \( x = 12 \)).

Problem (next, e.g., finding max/min of \( f(x) = (x + 3)^2 + 6 \))