Give an example of a quadratic function that has zeroes at x=2 and x=4, and that takes the value 6 when x=3.

An example of the quadratic function that has zeros at x = 2 and x = 4, and that takes the value 6 when x = 3 is  f(x) = -6·x² + 36·x  48

Step-by-step explanation:

The general form of a quadratic function is f(x) = a·x² + b·x + c

Which gives;

When x = 3, f(x) = 6 which gives;

6 = a·3² + b·3 + c  = a·9 + b·3 + c(1)

The zeros of the function are x = 2 and x = 4 which gives

0 = a·2² + b·2 + c = a·4 + b·2 + c(2)

0 = a·4² + b·4 + c = a·16 + b·4 + c(3)

Subtracting equation (2) from (3) gives;

a·16 + b·4 + c - (a·4 + b·2 + c)  = a·12 + b·2 = 0

b·2 = -a·12

b = -a·12/2 = -6·a

Substituting b = -6·a in equation (1) we have;

a·9 + b·3 + c = a·9 + (-6·a)·3 + c = 6

a·9 -18·a + c = 6

-9·a + c = 6

c = 6 + 9·a

Substituting b = -6·a and c = 6 + 9·a in equation (2), we get

0 = a·4 + b·2 + c = a·4 + (-6·a)·2 + 6 + 9·a = a·4 -12·a + 6 + 9·a

a + 6 = 0

a = - 6

b = -6·a = -6*(-6) = 36

c = 6 + 9·a = 6 + 9*(-6) = 6 - 54 = -48

The equation of the function is therefore, f(x) = -6·x² + 36·x  48.

(1, –1), (16, –4), (36, –6)

step-by-step explanation:

given that the function is

this can also be written as

we have to graph this function and check which list of points lie on the graph

see attachment for graph:

we find that the points lying are (1,-1) (16,-4) and (36,-6)

since square root is there we find that x can only be posiitve but y can be since square root both positive and negative.