To find the key points on the graph of the quadratic equation \( y = x^2 - 16x + 48 \), we will determine the vertex, x-intercepts, and y-intercept.
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Vertex: The vertex of a quadratic is given by the formula \( x = -\frac{b}{2a} \) where the equation is in the form \( y = ax^2 + bx + c \).
Here, \( a = 1 \), \( b = -16 \), and \( c = 48 \).
\[ x = -\frac{-16}{2 \cdot 1} = \frac{16}{2} = 8 \]
To find the y-coordinate of the vertex, substitute \( x = 8 \) back into the equation:
\[ y = (8)^2 - 16(8) + 48 = 64 - 128 + 48 = -16 \]
Therefore, the vertex is \( (8, -16) \).
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X-Intercepts: The x-intercepts occur when \( y = 0 \):
\[ 0 = x^2 - 16x + 48 \]
We can solve this quadratic using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, we have:
\[ x = \frac{16 \pm \sqrt{(-16)^2 - 4 \cdot 1 \cdot 48}}{2 \cdot 1} = \frac{16 \pm \sqrt{256 - 192}}{2} \]
\[ x = \frac{16 \pm \sqrt{64}}{2} = \frac{16 \pm 8}{2} \]
This gives us two results:
\[ x = \frac{24}{2} = 12 \quad \text{(larger x-intercept)} \]
\[ x = \frac{8}{2} = 4 \quad \text{(smaller x-intercept)} \]
So, the x-intercepts are \( (4, 0) \) and \( (12, 0) \).
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Y-Intercept: The y-intercept occurs when \( x = 0 \):
\[ y = (0)^2 - 16(0) + 48 = 48 \]
So, the y-intercept is \( (0, 48) \).
Summary of key points:
- Vertex: \( (8, -16) \)
- Smaller x-intercept: \( (4, 0) \)
- Larger x-intercept: \( (12, 0) \)
- Y-intercept: \( (0, 48) \)
Now, fill in the answers:
- vertex: \( (8, -16) \)
- (smaller) x-intercept: \( (4, 0) \)
- (larger) x-intercept: \( (12, 0) \)
- y-intercept: \( (0, 48) \)
1 answer