To determine which statement must be true given that [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex] are quadratic functions but [tex]\((f + g)(x)\)[/tex] produces the graph, let's analyze the given situation carefully.
Let's start by expressing [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex] in their general quadratic forms:
[tex]\[ f(x) = a_1x^2 + b_1x + c_1 \][/tex]
[tex]\[ g(x) = a_2x^2 + b_2x + c_2 \][/tex]
When these two functions are added together, the sum [tex]\( (f + g)(x) \)[/tex] will be:
[tex]\[ (f + g)(x) = (a_1x^2 + b_1x + c_1) + (a_2x^2 + b_2x + c_2) \][/tex]
Combining like terms:
[tex]\[ (f + g)(x) = (a_1 + a_2)x^2 + (b_1 + b_2)x + (c_1 + c_2) \][/tex]
The function [tex]\((f + g)(x)\)[/tex] will be quadratic if and only if the coefficient of [tex]\( x^2 \)[/tex] does not equal zero. Thus, for the resulting function [tex]\((f + g)(x)\)[/tex] to not produce a quadratic graph, the coefficient of [tex]\( x^2 \)[/tex] must be zero:
[tex]\[ a_1 + a_2 = 0 \][/tex]
This implies that the leading coefficients [tex]\( a_1 \)[/tex] and [tex]\( a_2 \)[/tex] of the quadratic functions [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex] must cancel each other out. Therefore, [tex]\( a_1 \)[/tex] and [tex]\( a_2 \)[/tex] must be opposites:
[tex]\[ a_1 = -a_2 \][/tex]
Thus, the correct statement is:
"The leading coefficients of [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex] are opposites."
