How do you find the x-intercepts, axis of symmetry, vertex, y-intercept, domain, and range of y=3x^2-6x+24?
The y-intercept is where x=0, so substituting 0 in for x, we get:
The x-intercept is where y=0, so:
We can factor out a 3 to get:
In order to solve for x^2 -2x+8, we use the Quadratic Formula, which gives our x-intercepts as:
x=1 +/- i*sq(7).
Axis of Symmetry:
To find the Axis of Symmetry, we can use the formula, x=-b/2a:
The Axis of Symmetry is the x-value of the Vertex, which we found to be 1.
To find the y-value of the Vertex, we substitute x=1 back into the original equation:
y=3(1)^2 -6(1) +24
y=21, so our Vertex is (1, 21).
The Domain is all real numbers.
A parabola opens up when a > 0 and opens down when a < 0. In this case, since a > 0, the Range is everything greater than the y-value of the Vertex, so:
The Range is, for all real numbers, y is greater than or equal to 21.