To find the key points on the graph of the given equation, we can analyze its form in vertex form: y = a(x - h)^2 + k.
The given equation, y = x^2 - 16x + 28, can be rewritten as y = (x^2 - 16x) + 28, which means a = 1, h = -16, and k = 28.
The vertex of the parabola is given by the coordinates (h, k), so the vertex here is (h, k) = (-16, 28).
To find the x-intercepts, we set y = 0 and solve for x. In this case, we need to solve the equation x^2 - 16x + 28 = 0.
Using the quadratic formula, x = (-b ± √(b^2 - 4ac)) / 2a, we can plug in the values a = 1, b = -16, and c = 28:
x = (-(-16) ± √((-16)^2 - 4(1)(28))) / (2 * 1)
Simplifying this expression, x = (16 ± √(256 - 112)) / 2
x = (16 ± √144) / 2
x = (16 ± 12) / 2
This gives us two solutions: x = (16 + 12) / 2 = 14, and x = (16 - 12) / 2 = 2. Therefore, the x-intercepts are at x = 14 and x = 2.
To find the y-intercept, we set x = 0 and solve for y. Plugging x = 0 into the equation y = x^2 - 16x + 28, we get y = 0^2 - 16(0) + 28 = 28.
Therefore, the y-intercept is at y = 28.
Summary of key points:
- Vertex: (-16, 28)
- X-intercepts: 14 and 2
- Y-intercept: 28
What are the key points on the graph of y=x2−16x+28%0D%0A%0D%0A=%0D%0A%0D%0A2%0D%0A−%0D%0A16%0D%0A%0D%0A+%0D%0A28%0D%0A ? Name the vertex, x-intercepts, and y-intercept.
1 answer