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2 months ago

Which statement is true regarding the graphed functions?

Mathematics

2 answers:

tester [12.3K]2 months ago

8 0

The right choice is $\boxed{\left(c\right)f\left(2\right)=0{\text{ and }}g\left({-2}\right)=0}$ .

Additional explanation:

Assign the function $f\left(x\right)$ as $y$ .

$\boxed{f\left(x\right)=y}$ ...... (1)

Define the function $g\left(x\right)$ as $z$ .

$\boxed{g\left(x\right)=z}$ ...... (2)

Replace 0 for $x$ in equation (1) to find $f\left(0\right)$ ; also, obtain $f\left(0\right)$ by checking the graph value of $y$ at $x=0$ .

$\boxed{f\left(0\right)=4}$

Insert 2 into $x$ in equation (1) to get $f\left(2\right)$ ; find $f(2)$ from the graph by locating $y$ at $x=2$ .

$\boxed{f\left(2\right)=0}$

Substitute $-2$ for $x$ in equation (2) to compute $g(-2)$ ; retrieve $g(-2)$ from the graph by finding $z$ at $x=-2$ .

$\boxed{g\left({-2}\right)=0}$

Check which option matches the results.

Option (a) shows $f(0)$ as 2, differing from the computed result, so it's incorrect.

Option (b) lists $f(0)$ as 4, not matching the calculated value, so it's incorrect.

Option (c) has $f(2)$ and $g(-2)$ both equal to 0, consistent with calculated values, so it's correct.

Option (d) shows $f(-2)$ as 0, but the graph reveals $f(-2)$ exceeds 12, making it incorrect.

Zina [12.3K]2 months ago

7 0

Let's examine each scenario to identify the correct statement.

Case 1) f(0) = 2 and g(–2) = 0

At x=0, locate f(0) on the graph: f(0) equals 4.

At x=-2, locate g(-2) on the graph: g(-2) equals 0.

Hence,

Case 1) statement is false.

Case 2) f(0) = 4 and g(–2) = 4

At x=0, f(0) from graph is 4.

At x=–2, g(–2) from graph is 0.

Therefore,

Case 2) statement is false.

Case 3) f(2) = 0 and g(–2) = 0

At x=2, from graph, f(2) equals 0.

At x=–2, g(–2) equals 0.

This means

Case 3) statement is true.

Case 4) f(–2) = 0 and g(–2) = 0

At x=–2, f(–2) appears greater than 12 on the graph.

At x=–2, g(–2) equals 0.

Thus,

Case 4) statement is false.

In conclusion,

The accurate statement is

f(2) = 0 and g(–2) = 0 — this statement holds true.

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