What are the key points on the graph of y = x^2 - 16x + 28? Name the vertex, x-intercepts, and y-intercept.
vertex: (__)
(smaller) x-intercept: (__)
(larger) x-intercept: (__)
y-intercept: (__)
To find the key points on the graph of the quadratic equation y = x^2 - 16x + 28, we can start by determining the vertex.
The equation is in the form y = ax^2 + bx + c, with a = 1, b = -16, and c = 28.
The x-coordinate of the vertex can be found using the formula -b/2a. In this case, it would be -(-16) / 2(1), which simplifies to 16/2 = 8.
To find the y-coordinate of the vertex, we substitute the x-coordinate (8) into the equation: y = (8)^2 - 16(8) + 28 = 64 - 128 + 28 = -36.
Therefore, the vertex is (8, -36).
Next, we can find the x-intercepts, which occur when y = 0.
0 = x^2 - 16x + 28
We can solve this quadratic equation by factoring or using the quadratic formula. In this case, the equation does not factor easily, so we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
Plugging in the values a = 1, b = -16, and c = 28:
x = (-(-16) ± √((-16)^2 - 4(1)(28))) / (2(1))
x = (16 ± √(256 - 112)) / 2
x = (16 ± √144) / 2
x = (16 ± 12) / 2
This gives us two solutions:
x1 = (16 + 12) / 2 = 28 / 2 = 14
x2 = (16 - 12) / 2 = 4 / 2 = 2
Thus, the smaller x-intercept is 2, and the larger x-intercept is 14.
To find the y-intercept, we set x = 0 in the equation:
y = (0)^2 - 16(0) + 28 = 0 - 0 + 28 = 28.
Therefore, the y-intercept is (0, 28).
In summary:
- Vertex: (8, -36)
- Smaller x-intercept: 2
- Larger x-intercept: 14
- Y-intercept: (0, 28)