To analyze the quadratic function \( y = x^2 - 16x + 48 \), we can find the vertex, x-intercepts, and y-intercept.
1. Vertex
The vertex of a quadratic function in the form \( y = ax^2 + bx + c \) can be found using the formula: \[ x = -\frac{b}{2a} \] For our function:
- \( a = 1 \)
- \( b = -16 \)
- \( c = 48 \)
Calculating the x-coordinate of the vertex: \[ x = -\frac{-16}{2 \cdot 1} = \frac{16}{2} = 8 \]
Now we find the y-coordinate by substituting \( x = 8 \) back into the equation: \[ y = (8)^2 - 16(8) + 48 = 64 - 128 + 48 = -16 \]
Therefore, the vertex is: \[ \text{Vertex} = (8, -16) \]
2. X-intercepts
To find the x-intercepts, we set \( y = 0 \): \[ 0 = x^2 - 16x + 48 \] We can solve this quadratic by factoring or using the quadratic formula. Let's first factor it: \[ x^2 - 16x + 48 = (x - 12)(x - 4) = 0 \] This gives us the x-intercepts: \[ x - 12 = 0 \quad \Rightarrow \quad x = 12 \ x - 4 = 0 \quad \Rightarrow \quad x = 4 \]
Thus, we have:
- Smaller x-intercept: \( (4, 0) \)
- Larger x-intercept: \( (12, 0) \)
3. Y-intercept
To find the y-intercept, we set \( x = 0 \): \[ y = (0)^2 - 16(0) + 48 = 48 \] So the y-intercept is: \[ \text{Y-intercept} = (0, 48) \]
Summary of Key Points
- Vertex: \( (8, -16) \)
- Smaller x-intercept: \( (4, 0) \)
- Larger x-intercept: \( (12, 0) \)
- Y-intercept: \( (0, 48) \)
Final Answers
- vertex: \( (8, -16) \)
- (smaller) x-intercept: \( (4, 0) \)
- (larger) x-intercept: \( (12, 0) \)
- y-intercept: \( (0, 48) \)