What are the key points on the graph of y=x^2−16x+28 ? Name the vertex, x-intercepts, and y-intercept.(2 points) vertex: () (smaller) x-intercept: () (larger) x-intercept: () y-intercept: ()
The key points on the graph of the quadratic equation y = x^2 - 16x + 28 are:
- Vertex: To find the vertex of the parabola, we can use the formula x = -b/2a. In this case, a = 1 and b = -16. Therefore, the x-coordinate of the vertex is x = -(-16)/(2*1) = 16/2 = 8. To find the y-coordinate, substitute the x-value into the equation: y = (8)^2 - 16(8) + 28 = 64 - 128 + 28 = -36. Therefore, the vertex is (8, -36).
- x-intercepts: To find the x-intercepts, we set y = 0 and solve for x. 0 = x^2 - 16x + 28. Since this quadratic doesn't factor easily, we can use the quadratic formula: x = (-b ± √(b^2 - 4ac))/(2a). In this case, a = 1, b = -16, and c = 28. Plugging these values into the formula, we get: x = (-(-16) ± √((-16)^2 - 4(1)(28)))/(2(1)) = (16 ± √(256 - 112))/2 = (16 ± √144)/2 = (16 ± 12)/2.
Simplifying further, we get:
x = (16 + 12)/2 = 28/2 = 14, and
x = (16 - 12)/2 = 4/2 = 2.
Therefore, the x-intercepts are (14, 0) and (2, 0).
- y-intercept: To find the y-intercept, we set x = 0 and solve for y. y = (0)^2 - 16(0) + 28 = 0 - 0 + 28 = 28. Therefore, the y-intercept is (0, 28).
The key points on the graph of y = x^2 - 16x + 28 are:
Vertex: (8, -36)
x-intercepts: (14, 0) and (2, 0)
y-intercept: (0, 28)