To analyze the quadratic function \( y = x^2 - 16x + 48 \), we will determine the vertex, x-intercepts, and y-intercept.
1. Finding the Vertex
The vertex of a quadratic \( y = ax^2 + bx + c \) can be found using the formula:
\[ x = -\frac{b}{2a} \]
For the function \( y = x^2 - 16x + 48 \):
- \( a = 1 \)
- \( b = -16 \)
Substituting into the formula:
\[ x = -\frac{-16}{2 \cdot 1} = \frac{16}{2} = 8 \]
To find the y-coordinate of the vertex, substitute \( x = 8 \) back into the equation:
\[ y = 8^2 - 16 \cdot 8 + 48 \] \[ y = 64 - 128 + 48 = -16 \]
Thus, the vertex is at:
\[ \text{vertex: } (8, -16) \]
2. Finding the x-intercepts
To find the x-intercepts, set \( y = 0 \):
\[ 0 = x^2 - 16x + 48 \]
This is a standard form quadratic equation. We can factor it:
\[ 0 = (x - 12)(x - 4) \]
Setting each factor to zero gives:
\[ x - 12 = 0 \quad \Rightarrow \quad x = 12 \] \[ x - 4 = 0 \quad \Rightarrow \quad x = 4 \]
Thus, the x-intercepts are:
\[ \text{smaller } x\text{-intercept: } (4, 0) \] \[ \text{larger } x\text{-intercept: } (12, 0) \]
3. Finding the y-intercept
To find the y-intercept, set \( x = 0 \):
\[ y = 0^2 - 16 \cdot 0 + 48 = 48 \]
Thus, 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: \( (0, 48) \)
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