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

*print*Print*list*Cite

### 3 Answers

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.**

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)**

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

At the point where the graph y = -3x^2+12x-21/2 crosses the x axis , the value of y = 0.

So -3x^2+12x-21/2 = 0.

We multiply by -2:

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

We divide by 3:

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

We know that the roots of ax^2+bx+c = 0 are : x1 = (-b+sqrt(b^2-4ac)}/2a or x2 = x1 = (-b-sqrt(b^2-4ac)}/2a.

So the rrots of 2x^2-8x+7 = 0 are:

x1 = {8+sqrt(8^2-4*2*7}/2*2 , or x2 = {8-sqrt(8^2-4*2*7}/2*2.

x1 = (4+ sqrt2)/2 = 2.71nearly , x2 = (4-sqrt2)/2 = 1.29nearly.

Therefore the y = -3x^2+12x-21/2 crosses the x axis at x = 2.71 and x = 1.29.