Question

I NEED HELP PLEASE! Anyone?...

Answer 1

To find the quadratic of the form ax^2+bx+c=y set up a system of equations with the points provided...

a+b+c=4, 0a+0b+c=3, a-b+c=-6  getting differences:

a+b=1, -a+b=9  and getting difference again:

2a=-8, so a=-4, making a+b=1 become:

-4+b=1, so b=5, making a+b+c=4 become:

-4+5+c=4, so c=3

thus the quadratic that produces those three points is:

y=-4x^2+5x+3

...

The vertex occurs when the velocity is equal to zero, the derivative of the quadratic...

dy/dx=-8x+5, dy/dx=0 when x=5/8=0.625

Then you can find y(5/8)=4.5625

So the vertex is (0.625, 4.5625)

Or you may wish to remember that the vertex will always be:

(-b/(2a), (4ac-b^2)/(4a))  (this can be derived from conservation of energy under constant acceleration or the vertex form of the quadratic)

The axis of symmetry will be the vertical line about the x-coordinate of the vertex, in this case:

x=0.625