To find the key points of the quadratic function \( y = x^2 - 16x + 48 \), we will find the vertex, x-intercepts, and y-intercept.
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Vertex: The vertex of a quadratic function in the form \( y = ax^2 + bx + c \) can be found using the formula: \[ x = -\frac{b}{2a} \] Here, \( a = 1 \) and \( b = -16 \).
\[ x = -\frac{-16}{2(1)} = \frac{16}{2} = 8 \]
Now, substituting \( x = 8 \) back into the function to find the y-coordinate of the vertex:
\[ y = (8)^2 - 16(8) + 48 = 64 - 128 + 48 = -16 \]
So, the vertex is \( (8, -16) \).
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X-intercepts: To find the x-intercepts, set \( y = 0 \): \[ 0 = x^2 - 16x + 48 \] Factoring the quadratic: \[ 0 = (x - 12)(x - 4) \]
Setting each factor equal to zero gives: \[ x - 12 = 0 \quad \Rightarrow \quad x = 12 \] \[ x - 4 = 0 \quad \Rightarrow \quad x = 4 \]
Thus, the smaller x-intercept is \( (4, 0) \) and the larger x-intercept is \( (12, 0) \).
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Y-intercept: To find the y-intercept, set \( x = 0 \): \[ y = 0^2 - 16(0) + 48 = 48 \] Thus, the y-intercept is \( (0, 48) \).
Putting it all together:
- Vertex: \( (8, -16) \)
- Smaller x-intercept: \( (4, 0) \)
- Larger x-intercept: \( (12, 0) \)
- Y-intercept: \( (0, 48) \)
Final answer:
- Vertex: \( (8, -16) \)
- Smaller x-intercept: \( (4, 0) \)
- Larger x-intercept: \( (12, 0) \)
- Y-intercept: \( (0, 48) \)
1 answer