Question

1. determine the vertex of each quadratic function in vertex form, then state whether it is a maximum or minimum. no graphing should be necessary!

a. ( a(x) = -\frac{1}{2}(x - 5)^2 + 4 )
b. ( b(x) = 2(x + 4)^2 )
c. ( c(x) = -(x - 5)^2 )
d. ( d(x) = x^2 + 8 )

Explanation:

Part a: \( a(x) = -\frac{1}{2}(x - 5)^2 + 4 \)

Step 1: Recall vertex form of a quadratic

The vertex form of a quadratic function is \( y = a(x - h)^2 + k \), where \((h, k)\) is the vertex.
For \( a(x) = -\frac{1}{2}(x - 5)^2 + 4 \), we compare with \( y = a(x - h)^2 + k \). Here, \( h = 5 \) and \( k = 4 \). So the vertex is \((5, 4)\).

Step 2: Determine maximum or minimum

The coefficient of \((x - 5)^2\) is \( -\frac{1}{2} \), which is negative. When the coefficient \( a \) in \( y = a(x - h)^2 + k \) is negative, the parabola opens downward, so the vertex is a maximum point.

Step 1: Rewrite in vertex form (if needed)

The function is already in vertex form \( y = a(x - h)^2 + k \), but here \( x + 4 = x - (-4) \), so \( h = -4 \) and \( k = 0 \) (since there's no constant term added). Thus, the vertex is \((-4, 0)\).

Step 2: Determine maximum or minimum

The coefficient of \((x + 4)^2\) is \( 2 \), which is positive. When \( a>0 \), the parabola opens upward, so the vertex is a minimum point.

Step 1: Identify vertex from vertex form

In \( y = a(x - h)^2 + k \) form, \( h = 5 \) and \( k = 0 \) (since there's no constant term added). So the vertex is \((5, 0)\).

Step 2: Determine maximum or minimum

The coefficient of \((x - 5)^2\) is \( -1 \), which is negative. So the parabola opens downward, and the vertex is a maximum point.

Answer:

(a): Vertex is \((5, 4)\), it is a maximum.

Part b: \( b(x) = 2(x + 4)^2 \)