Th graph of a quadratic function f(x)=hx²-12x+3k, where h and k are constants, has a minimum point.
(a)state the value of h, if -2＜h＜2 where h is an integer
(b)based on the answer in (a),find the value of k if the graph touches x-axis at one point

h = 1

k = 12.

Step-by-step explanation:

(A).

f(x ) = hx² - 12x + 3k

Convert to vertex form ( vertex form is a(x - b)^2 + c ):-

f(x) = h (x^2 - 12x/h) + 3k

=  h [ (x - 6/h)^2 - 36h^2 ] + 3k

= h(x - 6/h)^2 -  (36/h - 3k).

So the vertex is at x = 6/h.

h must be  positive because the function has a minimum value.

h < 2 so as its an integer it must be 1.

(B)

The vertex is at ( 6/h ,  -(36/h-3k) ).

If the graph touches the axis at one point then the vertex is (6/h , 0)

and - (36/h - 3k)  = 0

-36 = -3k

k = 12 (answer)

first one is a: renaming a fraction. second one is c: de-c

step-by-step explanation:

my explanation is that these were the answers on edg : )