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Answer:
[tex](-5)[/tex] and [tex]3[/tex].
Step-by-step explanation:
The [tex]x[/tex]-intercepts of a graph refer to points where the graph intersects the [tex]x\![/tex]-axis. These are [tex]x\!\![/tex] values for which [tex]f(x) = 0[/tex].
For example, since [tex]x = (-5)[/tex] is one of the [tex]x[/tex]-intercepts of [tex]y = f(x)[/tex], it must be true that [tex]f(-5) = 0[/tex]. Likewise, since [tex]3[/tex] is one of the [tex]x[/tex]-intercepts of [tex]y = f(x)\![/tex], [tex]f(3) = 0[/tex].
Since [tex]f(-5) = 0[/tex], the expression [tex]7\, f(-5)[/tex] would also evaluate to [tex]0[/tex] (that is, [tex]7\, f(-5) = (7)\, (0) = 0[/tex].) Thus, [tex]x = (-5)[/tex] would be an [tex]x[/tex]-intercept of the new graph [tex]y = 7\, f(x)[/tex].
Likewise, since [tex]f(3) = 0[/tex], [tex]7\, f(3) = (7)\, (0) = 0[/tex], such that [tex]x = 3[/tex] would also be an [tex]x[/tex]-intercept of the new graph [tex]y = 7\, f(x)[/tex].
No other points could be an [tex]x[/tex]-intercept of the new graph [tex]y = 7\, f(x)[/tex] without being an [tex]x\![/tex]-intercept of [tex]y = f(x)[/tex]. For example, assume that [tex]x = x_{0}[/tex] is an [tex]x[/tex]-intercept of [tex]y = 7\, f(x)[/tex] but not [tex]y = f(x)[/tex]. [tex]7\, f(x_{0}) = 0[/tex], such that [tex]f(x_{0}) = (1/7)\, (7\, f(x_{0})) = (1/7)\, (0) = 0[/tex]- contradiction.
Therefore, the [tex]x[/tex]-intercepts of the new graph would be [tex]x = (-5)[/tex] and [tex]x = 3[/tex].