To determine the value of [tex]\( x \)[/tex] at which the graphs intersect exactly at one point in the given equations, we need to find the points of intersection between the two curves and then apply the condition of unique intersection.
Given equations are:
[tex]\[ y = 2x^2 - 21x + 64 \][/tex]
[tex]\[ y = 3x + a \][/tex]
First, equate the two equations to find the points of intersection:
[tex]\[ 2x^2 - 21x + 64 = 3x + a \][/tex]
Rearranging the equation, we obtain:
[tex]\[ 2x^2 - 21x + 64 - 3x - a = 0 \][/tex]
[tex]\[ 2x^2 - 24x + (64 - a) = 0 \][/tex]
This is a quadratic equation of the form:
[tex]\[ 2x^2 - 24x + (64 - a) = 0 \][/tex]
For these equations to intersect at exactly one point, the quadratic equation should have exactly one solution. This occurs when the discriminant ([tex]\(\Delta\)[/tex]) of the quadratic equation is zero. The discriminant for a quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex] is given by:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
In our case, [tex]\( a = 2 \)[/tex], [tex]\( b = -24 \)[/tex], and [tex]\( c = 64 - a \)[/tex]. Plugging these values into the discriminant:
[tex]\[ \Delta = (-24)^2 - 4(2)(64 - a) \][/tex]
[tex]\[ \Delta = 576 - 8(64 - a) \][/tex]
[tex]\[ \Delta = 576 - 512 + 8a \][/tex]
[tex]\[ \Delta = 64 + 8a \][/tex]
Setting the discriminant to zero for exactly one solution:
[tex]\[ 64 + 8a = 0 \][/tex]
[tex]\[ 8a = -64 \][/tex]
[tex]\[ a = -8 \][/tex]
Now, substitute [tex]\( a = -8 \)[/tex] back into the quadratic equation to find the value of [tex]\( x \)[/tex]:
[tex]\[ 2x^2 - 24x + 72 = 0 \][/tex]
Divide through by 2 for simplicity:
[tex]\[ x^2 - 12x + 36 = 0 \][/tex]
[tex]\[ (x - 6)^2 = 0 \][/tex]
Thus,
[tex]\[ x = 6 \][/tex]
Therefore, the value of [tex]\( x \)[/tex] at which the graphs intersect exactly at one point is [tex]\( \boxed{6} \)[/tex].
