Y=x^2+8x+12
y-intercept?
X-intercept(s)?
Coordinates of the vertex?
Equation of axis of symmetry?
Maximum or minimum?
To find the y-intercept, plug in x=0 into the equation:
y = (0)^2 + 8(0) + 12 = 12
So, the y-intercept is at the point (0,12).
To find the x-intercepts, set y=0 and solve for x:
0 = x^2 + 8x + 12
0 = (x+2)(x+6)
x+2=0 or x+6=0
x=-2 or x=-6
So, the x-intercepts are at the points (-2,0) and (-6,0).
To find the coordinates of the vertex, use the formula for the x-coordinate of the vertex:
x = -b/2a = -8/(2*1) = -4
Now, plug in x=-4 into the equation to find the y-coordinate:
y = (-4)^2 + 8(-4) + 12 = 4 - 32 + 12 = -16
So, the coordinates of the vertex are (-4,-16).
The equation of the axis of symmetry is the vertical line passing through the vertex, which is x=-4.
Since the coefficient of x^2 is positive, the parabola opens upwards, so the vertex represents the minimum point of the parabola. So, the function has a minimum at (-4,-16).