What are the key points on the graph of y=x2−16x+48

? Name the vertex, x-intercepts, and y-intercept.(2 points)
vertex: (
)

(smaller) x-intercept: (
)

(larger) x-intercept: (
)

y-intercept: (
)

Remaining Attempts : 3
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1 answer

To find the key points of the quadratic function \( y = x^2 - 16x + 48 \), we need to complete the square or use the quadratic formula to determine the vertex and intercepts.

  1. Vertex: A quadratic can be expressed in the standard form \( y = ax^2 + bx + c \). For this function:

    • \( a = 1 \)
    • \( b = -16 \)
    • \( c = 48 \)

    The x-coordinate of the vertex is given by the formula: \[ x = -\frac{b}{2a} = -\frac{-16}{2 \cdot 1} = \frac{16}{2} = 8 \]

    Now, substitute \( x=8 \) back into the original equation to find the y-coordinate: \[ y = 8^2 - 16 \cdot 8 + 48 = 64 - 128 + 48 = -16 \]

    So, the vertex is: \[ \text{Vertex: } (8, -16) \]

  2. X-Intercepts: To find the x-intercepts, set \( y = 0 \): \[ 0 = x^2 - 16x + 48 \] We can factor this equation: \[ 0 = (x - 12)(x - 4) \] Setting each factor to zero gives us the x-intercepts: \[ x - 12 = 0 \quad \Rightarrow \quad x = 12 \] \[ x - 4 = 0 \quad \Rightarrow \quad x = 4 \]

    Therefore, the ((smaller) and (larger)) x-intercepts are: \[ \text{(smaller) x-intercept: } (4, 0) \] \[ \text{(larger) x-intercept: } (12, 0) \]

  3. Y-Intercept: To find the y-intercept, set \( x = 0 \): \[ y = 0^2 - 16 \cdot 0 + 48 = 48 \]

    Therefore, 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: \( (0, 48) \)

Fill in the values as follows:

  • vertex: \( (8, -16) \)
  • (smaller) x-intercept: \( (4, 0) \)
  • (larger) x-intercept: \( (12, 0) \)
  • y-intercept: \( (0, 48) \)