Questions: f(x)=x^2-4x+6 x=-(-4)/2(1)=0 f(x)=-x^2-8x-19

Transcript text: $f(x)=x^{2}-4 x+6 \quad x=-\frac{-4}{2(1)}=0$ $f(x)=-x^{2}-8 x-19$

Solution

Solution Steps

Step 1: Find the vertex of f(x) = x² - 4x + 6

The x-coordinate of the vertex is given by x = -b/2a. In this case, a = 1 and b = -4, so x = -(-4)/(2*1) = 2. Substituting x = 2 back into the function gives f(2) = (2)² - 4(2) + 6 = 4 - 8 + 6 = 2. Therefore, the vertex is at (2, 2). Since the coefficient of x² is positive, the parabola opens upwards, meaning the vertex is a minimum point.

Step 2: Find the vertex of f(x) = -x² - 8x - 19

The x-coordinate of the vertex is given by x = -b/2a. Here, a = -1 and b = -8, so x = -(-8)/(2*-1) = -4. Substituting x = -4 into the function yields f(-4) = -(-4)² - 8(-4) - 19 = -16 + 32 - 19 = -3. Therefore, the vertex is at (-4, -3). Since the coefficient of x² is negative, the parabola opens downwards, indicating that the vertex represents a maximum point.

Step 3: Summarize the results

For f(x) = x² - 4x + 6, the vertex is a minimum point at (2, 2). For f(x) = -x² - 8x - 19, the vertex is a maximum point at (-4, -3).