For the function y = 3x2 + 18x + 24 find the high and low x intercepts the y-intercept the vertex and the axis of symmetry
To find the x-intercepts, set y = 0 and solve for x:
0 = 3x^2 + 18x + 24
Divide the equation by 3:
0 = x^2 + 6x + 8
This quadratic equation can be factored as:
0 = (x + 4)(x + 2)
Setting each factor equal to zero gives the x-intercepts:
x + 4 = 0 -> x = -4
x + 2 = 0 -> x = -2
Therefore, the x-intercepts are -4 and -2.
To find the y-intercept, set x = 0:
y = 3(0)^2 + 18(0) + 24
y = 0 + 0 + 24
y = 24
Therefore, the y-intercept is 24.
The vertex of a parabola in the form y = ax^2 + bx + c can be found using the formula x = -b/2a. For this function, a = 3 and b = 18, so the x-coordinate of the vertex is:
x = -18 / (2 * 3)
x = -18 / 6
x = -3
To find the y-coordinate of the vertex, substitute this x-value back into the original function:
y = 3(-3)^2 + 18(-3) + 24
y = 3(9) - 54 + 24
y = 27 - 54 + 24
y = -3
Therefore, the vertex of the parabola is at (-3, -3).
The axis of symmetry is a vertical line that passes through the vertex. For this function, the axis of symmetry is x = -3.