If f(x)=x^2-6x+14 and g(x)=-x^2-20x-k, determine the value of k so that there is exactly one point of intersection between the two parabolas.

set them equal

x^2-6x+14=-x^2-20x-k
2x^2+14x+14+k=0

x= (-14+-sqrt (14^2 -56-4k))/4

so if there is one intersection,

196=56+k
solve for k.

To find the value of k so that the two parabolas, f(x) and g(x), have exactly one point of intersection, we need to set the two functions equal to each other and solve for x.

Setting f(x) = g(x), we have:

x^2 - 6x + 14 = -x^2 - 20x - k

Combining like terms, we get:

2x^2 - 14x - 14 = -k

Now, we want the parabolas to intersect at exactly one point. This means they will have a common x-coordinate, but different y-coordinates. In other words, the discriminant of the above equation (b^2 - 4ac) should be equal to zero.

In our case, a = 2, b = -14, and c = -14. Substituting these values into the discriminant formula, we have:

(-14)^2 - 4(2)(-14) = 196 + 112 = 308

To have exactly one point of intersection, the discriminant should be equal to zero. However, based on the calculation above, the discriminant is 308. This means there is no value of k that will result in exactly one point of intersection.