For a quadratic equation, the roots are `(-1,3)` and the minimum is 4 units, then the triangle formed by connecting the roots and the minimum value of y is

Options

A. right angle triangle
B. scalene triangle
C. equilateral triangle
D. isosceles triangle

Answer

To solve the problem step by step, we will analyze the given information about the quadratic equation and the triangle formed by its roots and the minimum value of y.

### Step 1: Identify the Roots
The roots of the quadratic equation are given as \( x_1 = -1 \) and \( x_2 = 3 \).

### Step 2: Plot the Roots on a Graph
On a Cartesian plane, plot the points corresponding to the roots:
- Point A at \( (-1, 0) \)
- Point B at \( (3, 0) \)

### Step 3: Identify the Minimum Value
The minimum value of the quadratic function is given as \( y = 4 \). This means there is a point C at \( (x, 4) \) where the vertex of the parabola is located. Since the vertex lies directly above the midpoint of the roots, we find the midpoint \( x \) as follows:
\[ x = \frac{x_1 + x_2}{2} = \frac{-1 + 3}{2} = 1 \]
Thus, point C is at \( (1, 4) \).

### Step 4: Determine the Vertices of the Triangle
Now we have the vertices of the triangle:
- A: \( (-1, 0) \)
- B: \( (3, 0) \)
- C: \( (1, 4) \)

### Step 5: Calculate the Lengths of the Sides of the Triangle
1. **Length of AB** (base):
\[ AB = |x_2 - x_1| = |3 - (-1)| = 4 \text{ units} \]

2. **Length of AC**:
\[ AC = \sqrt{(1 - (-1))^2 + (4 - 0)^2} = \sqrt{(1 + 1)^2 + 4^2} = \sqrt{2^2 + 4^2} = \sqrt{4 + 16} = \sqrt{20} \text{ units} \]

3. **Length of BC**:
\[ BC = \sqrt{(3 - 1)^2 + (0 - 4)^2} = \sqrt{(2)^2 + (-4)^2} = \sqrt{4 + 16} = \sqrt{20} \text{ units} \]

### Step 6: Analyze the Triangle
Now we have the lengths of the sides of the triangle:
- \( AB = 4 \)
- \( AC = \sqrt{20} \)
- \( BC = \sqrt{20} \)

Since \( AC \) and \( BC \) are equal, the triangle formed is an isosceles triangle.

### Conclusion
The triangle formed by connecting the roots and the minimum value of y is an **isosceles triangle**.

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