For which values of x does the graph of the equation y=-3x^2 +12x -21/2 cross the x-axis.

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The curve y=-3x^2 +12x -21/2 crosses the x axis when y = 0.

We can determine y = 0 by equating -3x^2 +12x -21/2 to zero

=> -3x^2 +12x -21/2 = 0

multiply by 2

=> -6x^2 + 24x - 21 = 0

divide by -3

=> 2x^2 - 8x + 7 = 0

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

=> [ 8 + sqrt (64 - 56)]/4

=> 2 + (sqrt 2)/2

x2 = [ 8 - sqrt (64 - 56)]/4

=> 2 - (sqrt 2)/2

So the graph crosses the x axis at x equal to 2 + (sqrt 2)/2 and 2 - (sqrt 2)/2.

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We have the curve :

y= -3x^2 + 12x - 21/2

We need to determine the points where the curve meets the x-axis.

When y meets the x-axis, then the values of y will be zero.

Then, we will substitute with y= 0 and determine the x-coordinates.

==> -3x^2 + 12x - 21/2 = 0

==> First we will multiply by 2.

==> -6x^2 + 24x - 21 = 0

Now we will divide by 3.

==> -2x^2 + 8x -7 = 0

==> x1= (-8 + sqrt(64-4*-2*-7) / -4

           = (-8 + sqrt(8) / -4 = (-8+2sqrt2)/-4

==> x1= 2 - sqrt2 /2

==> x2= 2 + sqrt2 / 2

Then, the curve y meets the x-axis at the points:

( 2-sqrt2 /2 , 0) and ( 2+sqrt2 /2 , 0)

 

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