Answer:
[tex]k = 1[/tex].
[tex]h = 5[/tex].
After reflection: [tex]f(x) = x^{2} - 6\, x - 5[/tex].
Step-by-step explanation:
If the vertex of a quadratic function is at [tex](x_{0},\, y_{0})[/tex], the vertex form equation of this parabola would be:
[tex]f(x) = a\, (x - x_{0})^{2} + y_{0}[/tex]
For some nonzero constant [tex]a[/tex].
In this question, it is given that the vertex of the quadratic equation [tex]f(x)[/tex] is at [tex](3,\, 14)[/tex]. Hence, there would exist a non-zero constant [tex]a[/tex] such that:
[tex]f(x) = a\, (x - 3)^{2} + 14[/tex].
Rearrange this equation to match the format [tex]f(x) = k\, x^{2} + 6\, x + h[/tex]:
[tex]f(x) = a\, x^{2} + (- 6\, a)\, x + (9\, a + 14)[/tex].
For the two expressions to be equivalent, [tex]k = a[/tex], [tex]6 = (-6\, a)[/tex], and [tex]h = 9\, a + 14[/tex]. Solve this system of equations to obtain:
[tex]a = (-1)[/tex].
[tex]k = (-1)[/tex].
[tex]h = 5[/tex].
Hence, [tex]f(x) = -x^{2} + 6\, x + 5[/tex].
When the graph of a function is reflected with respect to the horizontal [tex]x[/tex]-axis, the new function value would be the opposite of the original one:
[tex](-f(x)) = -\left(-x^{2} + 6\, x + 5\right) = x^{2} - 6\, x - 5[/tex].
