To find the key points on the graph of the quadratic function \( y = x^2 - 16x + 48 \), we can determine the vertex, x-intercepts, and y-intercept.
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Vertex: The vertex of a parabola given in the standard form \( y = ax^2 + bx + c \) can be found using the formula for the x-coordinate of the vertex: \[ x = -\frac{b}{2a} \] For this function:
- \( a = 1 \)
- \( b = -16 \)
Plugging in the values: \[ x = -\frac{-16}{2 \cdot 1} = \frac{16}{2} = 8 \] Now, substitute \( x = 8 \) back into the equation to find the corresponding y-coordinate: \[ y = (8)^2 - 16(8) + 48 = 64 - 128 + 48 = -16 \] Thus, the vertex is at \( (8, -16) \).
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X-Intercepts: The x-intercepts occur where \( y = 0 \). We set the equation equal to zero and solve for \( x \): \[ 0 = x^2 - 16x + 48 \] Factoring the quadratic: \[ 0 = (x - 6)(x - 8) \] Setting each factor to zero gives: \[ x - 6 = 0 \quad \Rightarrow \quad x = 6 \] \[ x - 8 = 0 \quad \Rightarrow \quad x = 8 \] So, the x-intercepts are \( (6, 0) \) and \( (8, 0) \).
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Y-Intercept: The y-intercept occurs where \( x = 0 \). We evaluate the function at \( x = 0 \): \[ y = 0^2 - 16(0) + 48 = 48 \] Thus, the y-intercept is \( (0, 48) \).
Summary of Key Points:
- Vertex: \( (8, -16) \)
- X-Intercepts: \( (6, 0) \) and \( (8, 0) \)
- Y-Intercept: \( (0, 48) \)