Question

What are the key points on the graph of y = x^2 - 16x + 48? Name the vertex, x-intercepts, and y-intercept.

Answer 1

good but I also need the smaller x intercept and the larger x intercept

Answer 2

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.

1. **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 \( (8, -16) \).

2. **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 \( (8, 0) \).

3. **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**: \( (0, 48) \)

Answer 3

Certainly! From the previous analysis, we determined the x-intercepts for the quadratic equation \( y = x^2 - 16x + 48 \) as follows:

1. The x-intercepts were found by solving the equation \( 0 = (x - 6)(x - 8) \), yielding two solutions:
- \( x = 6 \)
- \( x = 8 \)

To clarify:
- The **smaller x-intercept** is \( (6, 0) \).
- The **larger x-intercept** is \( (8, 0) \).

### Summary of X-Intercepts:
- **Smaller x-intercept**: \( (6, 0) \)
- **Larger x-intercept**: \( (8, 0) \)