For each relation, find the y-intercept and state if there is a maximum or a minimum. Relations: y=x^2-25 , y=-4x^2-16 , (c)y=0.25x^2+2x+8. What is the y intercept and maximum or minimum of those equations?

Answer 1

1. For y = x^2 - 25:
- The y-intercept occurs when x = 0:
y = 0^2 - 25 = -25
- Since the coefficient of x^2 is positive (1), the parabola opens upwards. Therefore, the vertex is the minimum point.
- The minimum occurs at the vertex, where x = 0:
y = 0^2 - 25 = -25
- The y-intercept is -25 and the parabola has a minimum at y = -25.

2. For y = -4x^2 - 16:
- The y-intercept occurs when x = 0:
y = -4(0)^2 - 16 = -16
- Since the coefficient of x^2 is negative (-4), the parabola opens downwards. Therefore, the vertex is the maximum point.
- The maximum occurs at the vertex, where x = 0:
y = -4(0)^2 - 16 = -16
- The y-intercept is -16 and the parabola has a maximum at y = -16.

3. For y = 0.25x^2 + 2x + 8:
- The y-intercept occurs when x = 0:
y = 0.25(0)^2 + 2(0) + 8 = 8
- Since the coefficient of x^2 is positive (0.25), the parabola opens upwards. Therefore, the vertex is the minimum point.
- To find the x-coordinate of the vertex, we use x = -b/(2a):
x = -2/(2*0.25) = -2/(0.5) = -4
- Substitute x = -4 back into the equation to find the y-coordinate:
y = 0.25(-4)^2 + 2(-4) + 8 = 4 + (-8) + 8 = 4
- The vertex occurs at x = -4, y = 4.
- The y-intercept is 8 and the vertex is at x = -4, y = 4 making it the minimum.