What are the key points on the graph of y=x2−16x+48y=x2−16x+48? Name the vertex, x-intercepts, and y-intercept.(2 points) vertex: () (smaller) x-intercept: () (larger) x-intercept: () y-intercept: (

1 answer

To analyze the quadratic function \( y = x^2 - 16x + 48 \), we will determine the vertex, x-intercepts, and y-intercept.

1. Vertex

The vertex of a parabola defined by the equation \( y = ax^2 + bx + c \) can be found using the formula \( x = -\frac{b}{2a} \).

Here, \( a = 1 \), \( b = -16 \), and \( c = 48 \).

\[ x = -\frac{-16}{2 \cdot 1} = \frac{16}{2} = 8 \]

Now we substitute \( x = 8 \) back into the equation to find the y-coordinate:

\[ y = (8)^2 - 16(8) + 48 = 64 - 128 + 48 = -16 \]

Thus, 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 quadratic equation:

\[ 0 = (x - 12)(x - 4) \]

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: \[ \text{Smaller x-intercept: } (4, 0), \quad \text{Larger x-intercept: } (12, 0) \]

3. Y-Intercept

To find the y-intercept, set \( x = 0 \):

\[ y = (0)^2 - 16(0) + 48 = 48 \]

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)