SOLUTION: This problem is a porabola equation. i really need help.
The question is, : (A.) tell whether the graph opens up or down. (b) find the coordinatnts of the vertex. (c) Write an eqq
Question 448426: This problem is a porabola equation. i really need help.
The question is, : (A.) tell whether the graph opens up or down. (b) find the coordinatnts of the vertex. (c) Write an eqquation of the axis of symmetry.(D) Find the roots of the equation. and graph the function. The equation is: 2x^2-4x-16=y Answer by ewatrrr(24785) (Show Source):
Hi
y= 2x^2-4x -16
Note: the vertex form of a parabola, where(h,k) is the vertex
y= 2x^2-4x -16 |Completing the square
y = 2[(x-1)^2 - 1] -16
y = 2(x-1)^2 - 2 -16
y = 2(x-1)^2 -18 Vertex(1,-18) a = 2 > 0 parabola opens upward
Line of symmetry is x = 1
x-intercepts when y = 0 are also the solutions for the quadratic equation
2(x-1)^2 = 18
x-1 = ±
x = 1 ± 3 x = 4 or x = -2
