If f(x) and g(x) are quadratic functions but (f + g)(x) produces the graph below, which statement must be true? On a coordinate plane, a straight line with negative slope represents (f + g) (x). It goes through points (negative 3, 4), (0, 1), and (1, 0).

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Answer:

You did not write the options, but let's solve it in a general way.

f(x) and g(x) are quadratic functions, this means that:

f(x) = a*x^2 + b*x + c

g(x) = d*x^2 + e*x + f

now, (g + f)(x) is a linear function that goes trough the points (-3, 4), (0,1 ) and (1,0)

(g + f)(x) = s*x + k

We can find the slope of this linear function as:

S = (y2 - y1)/(x2 - x1)

i will use the first two points.

S = (1 - 4)/(0 - (-3) = -3/3 = -1

so we have: ( f + h)(x) = -1*x + k

and we can find k using one of the points, for example (0, 1)

1 = -1*0 + k

b = 1.

so we have: (f + g)(x) = -1*x + 1.

Now, (f+g)(x) = (a + d)*x^2 + (b + e)*x + (c + f) = -1*x + 1.

So we must have that:

a = -d

b + e = -1

c + f = 1.

Answer:

A. The leading coefficients of f(x) and g(x) are opposites.Step-by-step explanation: