Y=x^2+2x-80.
y-intercept?
X-intercept(s)?
Coordinates of the vertex?
Equation of axis of symmetry?
Maximum or minimum?
To find the y-intercept, we let x = 0 and solve for y:
y = (0)^2 + 2(0) - 80
y = -80
So the y-intercept is at (0, -80).
To find the x-intercepts, we set y = 0 and solve for x:
0 = x^2 + 2x - 80
0 = (x + 10)(x - 8)
So the x-intercepts are at (-10, 0) and (8, 0).
To find the coordinates of the vertex, we can use the formula x = -b/2a:
x = -2 / (2*1)
x = -1
Substitute x = -1 into the equation to find y:
y = (-1)^2 + 2(-1) - 80
y = -79
So the coordinates of the vertex are (-1, -79).
The equation of the axis of symmetry is x = -b/(2a), which in this case is x = -2/(2*1) = -1.
Since the coefficient of x^2 is positive, the parabola opens upwards and the vertex is the minimum point. So the minimum value of y is -79.