If k>0 and the graph y = x^2 - kx + 16 is tangent to the x-axis, what is the value of k?

Question

I'm not sure about this question either:

f(x) = -x^2 + k, what is the value of k if the endpoints of the parabola is (-2,0) and (2,0).

Answer 1

For y = x² - kx + 16 to be tangent to the x-axis, it has to be a perfect square, in this case (x-4)², which makes k = 8.

If f(x) = -x² + k and (-2,0) and (2,0) are on the graph (x-axis intersection points, not endpoints), then k is obviously 4, since 0 = -(-2)² + k, 0 = -4 + k, 4 = k, and likewise for (2,0).

Answer 2

First problem: assume that k = 8, that being twice the square root of 16, and verify that it solves the equation if x = 4. It does, so we're done.

Second problem. Obviously, we have symmetry about the y-axis, where x = 0. If the parabola passes through (2, 0) then we have
-(2)^2 + k = 0, so k = 4.

Answer 3

y = x^2 - kx + 16

then parabola must have a double equal root on the x-axis in order to be tangent to the x-axis.

take the discriminant and set it equal to zero.

D = sqrt(b^2 - 4ac)
0 = sqrt(k^2 - 64)

square both sides

0 = k^2 - 64
-k^2 = - 64
k^2 = 64
k = + or - 8

try graphing it, you'll see why k is + or - 8.


2)

F(x) = -x^2 + k

sub in one pair of coordinates to solve for k
let's use (2,0)

0 = -(2)^2 + k
k = 4

F(x) = -x^2 + 4
is the equation

Answer 4

1)
Since the graph is tangent to the x-axis, we have y = 0 and y' = 0.
y' = 2x-k = 0 => k=2x
y = x^2-2x*x+16 = 0
x = ±4
k = 2x = 8 since k>0

2)
Since f(x) = -x^2+k
k = f(x)+x^2 = 0+2^2 = 4