To analyze the function \( f(x) = 2x^2 + 16x - 10 \), we can follow the steps outlined in your questions.
### (a) Vertex
The vertex of a quadratic function in the form \( f(x) = ax^2 + bx + c \) can be found using the formula:
\[
x = -\frac{b}{2a}
\]
Here, \( a = 2 \) and \( b = 16 \).
Calculating the x-coordinate of the vertex:
\[
x = -\frac{16}{2 \cdot 2} = -\frac{16}{4} = -4
\]
Now, substituting \( x = -4 \) back into the function to find the y-coordinate:
\[
f(-4) = 2(-4)^2 + 16(-4) - 10 = 2(16) - 64 - 10 = 32 - 64 - 10 = -42
\]
Thus, the vertex is at the point:
\[
\boxed{(-4, -42)}
\]
### (b) Maximum or Minimum Value
Since the coefficient of \( x^2 \) (which is \( a = 2 \)) is positive, the parabola opens upward. Therefore, it has a minimum value at the vertex.
The minimum value of the function is:
\[
\boxed{-42}
\]
### (c) Range of \( f(x) \)
Since the parabola opens upward and the minimum value is at \( -42 \), the range of \( f(x) \) is:
\[
[-42, \infty)
\]
So, the answer in interval notation is:
\[
\boxed{[-42, \infty)}
\]
### (d) Increasing Interval
The function \( f(x) \) is increasing when \( x \) is greater than the x-coordinate of the vertex. Since the vertex is at \( x = -4 \), the function is increasing for:
\[
(-4, \infty)
\]
Thus, the answer in interval notation is:
\[
\boxed{(-4, \infty)}
\]
In summary:
- (a) Vertex: \((-4, -42)\)
- (b) Minimum value: \(-42\)
- (c) Range: \([-42, \infty)\)
- (d) Increasing on: \((-4, \infty)\)
(a) Vertex: \((-4, -42)\)
(b) Minimum value: \(-42\)
(c) Range: \([-42, \infty)\)
(d) Increasing on: \((-4, \infty)\)