Questions: MATH 108061College Algebra USCAFall 2024 Quiz: Practice Problems 2 (Part of grade calculat (Copy) Question list Given f(x) = 2(x + 1)^2 + 7 a. State the coordinates of the vertex. b. Identify the interval on which f(x) is increasing. c. State the range. d. State the axis of symmetry. Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 Question 10 October 08 26

Solution Steps

a. The function \( f(x) = 2(x + 1)^2 + 7 \) is in vertex form \( a(x-h)^2 + k \), where \((h, k)\) is the vertex. Here, \( h = -1 \) and \( k = 7 \), so the vertex is \((-1, 7)\).

b. Since the coefficient of the squared term is positive, the parabola opens upwards. The function is increasing on the interval to the right of the vertex, which is \((-1, \infty)\).

c. The range of the function is determined by the vertex and the direction the parabola opens. Since it opens upwards and the vertex is at its minimum point, the range is \([7, \infty)\).

The vertex of the function \( f(x) = 2(x + 1)^2 + 7 \) is given by the coordinates \( (h, k) \). Here, \( h = -1 \) and \( k = 7 \). Therefore, the vertex is: \[ \boxed{(-1, 7)} \]

Since the parabola opens upwards (as the coefficient of the squared term is positive), the function \( f(x) \) is increasing for all \( x \) values greater than the x-coordinate of the vertex. Thus, the interval on which \( f(x) \) is increasing is: \[ \boxed{(-1, \infty)} \]

The range of the function is determined by the minimum value at the vertex and the direction in which the parabola opens. Since the vertex is at \( k = 7 \) and the parabola opens upwards, the range of \( f(x) \) is: \[ \boxed{[7, \infty)} \]

Final Answer