Questions: Given the following function, (a) find the vertex; (b) determine whether there is a maximum or a minimum value, and find the value; (c) find the range; and (d) find the intervals on which the function is increasing and the intervals on which the function is decreasing. f(x)=4 x^2+24 x-28

Given the following function, (a) find the vertex; (b) determine whether there is a maximum or a minimum value, and find the value; (c) find the range; and (d) find the intervals on which the function is increasing and the intervals on which the function is decreasing.
f(x)=4 x^2+24 x-28
Transcript text: Given the following function, (a) find the vertex; (b) determine whether there is a maximum or a minimum value, and find the value; (c) find the range; and (d) find the intervals on which the function is increasing and the intervals on which the function is decreasing. \[ f(x)=4 x^{2}+24 x-28 \]
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Solution

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Solution Steps

To solve the given quadratic function \( f(x) = 4x^2 + 24x - 28 \):

(a) Find the vertex of the quadratic function. The vertex form of a quadratic function is given by \( f(x) = a(x-h)^2 + k \), where \((h, k)\) is the vertex. For a quadratic function in standard form \( ax^2 + bx + c \), the vertex can be found using the formula \( h = -\frac{b}{2a} \) and then substituting \( h \) back into the function to find \( k \).

(b) Determine whether the function has a maximum or minimum value. Since the coefficient of \( x^2 \) (which is 4) is positive, the parabola opens upwards, indicating a minimum value at the vertex.

(c) Find the range of the function. Since the parabola opens upwards and the minimum value is at the vertex, the range will be from the minimum value to infinity.

(d) Find the intervals on which the function is increasing and decreasing. The function will be decreasing on the interval to the left of the vertex and increasing on the interval to the right of the vertex.

Step 1: Find the Vertex

To find the vertex of the quadratic function \( f(x) = 4x^2 + 24x - 28 \), we use the formula for the vertex \( (h, k) \), where \( h = -\frac{b}{2a} \). Here, \( a = 4 \) and \( b = 24 \).

Calculating \( h \): \[ h = -\frac{24}{2 \cdot 4} = -3 \]

Next, we substitute \( h \) back into the function to find \( k \): \[ k = f(-3) = 4(-3)^2 + 24(-3) - 28 = -64 \]

Thus, the vertex is: \[ \text{Vertex} = (-3, -64) \]

Step 2: Determine Maximum or Minimum Value

Since the coefficient \( a = 4 \) is positive, the parabola opens upwards. Therefore, the function has a minimum value at the vertex. The minimum value is: \[ \text{Minimum Value} = -64 \]

Step 3: Find the Range

The range of the function is determined by the minimum value, which is the lowest point on the graph. Since the parabola opens upwards, the range is: \[ \text{Range} = [-64, \infty) \]

Step 4: Intervals of Increasing and Decreasing

The function is decreasing on the interval to the left of the vertex and increasing on the interval to the right of the vertex. Therefore, the intervals are: \[ \text{Decreasing Interval} = (-\infty, -3) \] \[ \text{Increasing Interval} = (-3, \infty) \]

Final Answer

  • Vertex: \(\boxed{(-3, -64)}\)
  • Minimum Value: \(\boxed{-64}\)
  • Range: \(\boxed{[-64, \infty)}\)
  • Decreasing Interval: \(\boxed{(-\infty, -3)}\)
  • Increasing Interval: \(\boxed{(-3, \infty)}\)
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