Maths help needed urgently..?

Question

the zeros of any quadratic function y=ax+bx+c , a≠0, are given by the quadratic formula X = -b + √b^2 -- 4ac

2a

show that the turning point of parabola is given by the coordinates

(-b , 4ac-b2 )

(2a 4a )

[their is a line where the new line began..]

thats the question and i am totally clueless.. could u work it out and show working too so i now what to do in the test?>?

thanks.

Answer 1

The turning point is on the axis of symmetry. When a parabola written in standard form y = a(x+h)² + k we know the axis of symmetry is located at x=-h.

Expand y = a(x+h)² + k

y = a(x² + 2xh + h²) + k

y = ax² + 2xha + h²a + k =

y = ax² + bx + c if b=2ha and c=(h²a + k)

Since the axis of symmetry is at x=-h, and h=b/2a, x= -b/2a.

To find the y coordinate, plug in x=-b/2a

y = a(-b/2a)² + b(-b/2a) + c

y = b²/4a - b²/2a + c

y = (4ac - b²)/4a

The coordinates of the turning point are [-b/2a, (4ac - b²)/4a]

Answer 2

The meaning of "turning point" is not clear.  If you mean the minimum value of the function, a bit of differential calculus serves to find it: dF/dx = 2ax + b, equate to zero to get x = -b/(2a). Plug this back into the original formula to get the corresponding y coordinate.  

Answer 3

The turning point is midway between the zeros of the parabola

If you have two points, say x1 and x2, then the midpoint is given by (x1 + x2)/2.

In this case you have:

x1 = [-b + SQRT(b^2 - 4ac)]/2a

x2 = [-b - SQRT(b^2 - 4ac)]/2a

The midpoint is then

x = [(-b/2a) + (-b/2a)]/2 = -b/2a

because the + and - SQRT terms are identical and they cancel out.

That solves it for x. To find y you just put it into the equation.

y = ax^2 + bx + c

y = a[b^2/4a^2] + b(-b/2a) + c

y = b^2/4a - b^2/2a + c

y = (b^2 - 2b^2 + 4ac)/4a

y = (4ac - b^2)/4a

And that is the y value.