The difference of 2 squares...?

Hi,

0 = 6x² + 12x - 18

0 = 6(x² + 2x - 3)

0 = 6(x + 3)(x - 1)

x + 3 = 0
x = -3 <==ANSWER for your equation

x - 1 = 0
x = 1 <==ANSWER for your equation

These "answers" are the roots or x intercepts of the graph.

If you want the "turning point" of the graph, that is its vertex. The vertex of a parabola is always on the axis of symmetry. The equation of this axis is found by x = -b/(2a). For this equation, the axis is:

x = -12/(2*6)
x = -1

That means the vertex is at (-1, ?). To find the y value, substitute -1 into the equation y = 6x² + 12x - 18 and solve for y.

y = 6x² + 12x - 18
y = 6(-1)² + 12(-1) - 18
y = -24

So the vertex is (-1, -24) <==ANSWER
The "turning point" is at (-1,-24).

I hope that helps!! :-)

1) -2^2 the square has precidence over the negative so it equals -4. In (-2)^2 the negative has precidence so the answer is 4. 2) To subtract a polynomial h(x) from g(x) multiply h(x) by (-1) and add like terms. 3) Again, watch the parentheses. x^2 + y^2 is simply that. (x+y)^2 can be simplified to x^2 +2xy + y^2 using the FOIL method.

0 = 6x^2 + 12x - 18
6(x^2 + 2x - 3) = 0
x^2 + 2x - 3 = 0/6
x^2 + 3x - x - 3 = 0
(x^2 + 3x) - (x + 3) = 0
x(x + 3) - 1(x + 3) = 0
(x + 3)(x - 1) = 0

x + 3 = 0
x = -3

x - 1 = 0
x = 1

∴ x = -3 , 1

f (x) = 6x² + 12x - 18
f (x) = 6 (x² + 2x - 3)
f (x) = 6 (x² + 2x + 1 - 4)
f (x) = 6 (x + 1) ² - 24
Turning point (-1,-24)

OR by Calculus :-

f ` (x) = 12 x + 12 = 0 for turning point
x = - 1
f(-1) = 6 - 12 - 18 = - 24
Turning point (-1.-24)

Your approach appears to be faulty. Proceed as follows:
Let f(x)= 6x^2 + 12x -18
To find the turning points,we have to find out maxima and minima of the expression. It is like this:
6x² + 12x -18 = 6(x² + 2x−3) =6{(x² + 2x + 1) −4)}
=6{(x + 1)² −2²)}
This has a a minimum value of −64 when x = −1
This expression has one turning point at x = −1

For a function: y = f(x), a stationary point / turning point is a point on the function graph where the gradient of the function is zero. I suppose your equation is the gradient equation of the graph / function. Your equation is actually quadratic in x ( I don't understand why you say you need to work out the difference of two squares of this equation.) Solve for x as follows :

Divide the equation by 6 through out resulting in x^2 + 2x -3 = 0
which is same as ( x -1 ) * ( x + 3 ) = 0. Your curve has two turning points one at x = +1 and another at x = -3. OK

= 6 ( x^2 + 2x - 3) = 0

x^2 -x +3x -3 = 0

x( x-1) +3(x-1) = 0

(x-1)(x+3)=0

x-1=0 , x+3 =0

x = 1 x = -3