To analyze the quadratic function \( y = x^2 - 16x + 28 \), we will find the vertex, x-intercepts (roots), and y-intercept.
Step 1: Finding the Vertex
The vertex of a quadratic function in the form \( y = ax^2 + bx + c \) can be found using the formula for the x-coordinate of the vertex:
\[ x = -\frac{b}{2a} \]
In this case, \( a = 1 \) and \( b = -16 \). Plugging in these values:
\[ x = -\frac{-16}{2 \cdot 1} = \frac{16}{2} = 8 \]
Now, we substitute \( x = 8 \) into the original equation to find the y-coordinate:
\[ y = (8)^2 - 16(8) + 28 = 64 - 128 + 28 = -36 \]
So, the vertex is:
\[ \text{vertex: } (8, -36) \]
Step 2: Finding the x-intercepts
To find the x-intercepts, we set \( y = 0 \):
\[ 0 = x^2 - 16x + 28 \]
This is a quadratic equation, and we can solve it using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
Here, \( a = 1 \), \( b = -16 \), and \( c = 28 \). We calculate the discriminant:
\[ b^2 - 4ac = (-16)^2 - 4 \cdot 1 \cdot 28 = 256 - 112 = 144 \]
Now, apply the quadratic formula:
\[ x = \frac{-(-16) \pm \sqrt{144}}{2 \cdot 1} = \frac{16 \pm 12}{2} \]
Calculating the two values:
- \( x = \frac{16 + 12}{2} = \frac{28}{2} = 14 \)
- \( x = \frac{16 - 12}{2} = \frac{4}{2} = 2 \)
Thus, the x-intercepts are:
\[ \text{smaller x-intercept: } (2, 0) \] \[ \text{larger x-intercept: } (14, 0) \]
Step 3: Finding the y-intercept
The y-intercept occurs when \( x = 0 \):
\[ y = (0)^2 - 16(0) + 28 = 28 \]
Thus, the y-intercept is:
\[ \text{y-intercept: } (0, 28) \]
Summary of Key Points
- Vertex: \( (8, -36) \)
- Smaller x-intercept: \( (2, 0) \)
- Larger x-intercept: \( (14, 0) \)
- y-intercept: \( (0, 28) \)