Match the key aspect of a function's graph with its meaning.
f(x) > 0
intervals of the domain where the
graph is above the x-axis
f(x) < 0
location on graph where input is zero
x-intercept
location on graph where output is
zero
y-intercept
intervals of the domain where the
graph is below the x-axis

Part 1) Intervals of the domain where the  graph is above the x-axis (f(x) > 0)

Part 2) location on graph where input is zero  (y-intercept)

Part 3) location on graph where output is zero  (x-intercept)

Part 4) Intervals of the domain where the  graph is below the x-axis (f(x) < 0)

Step-by-step explanation:

Verify each case

Part 1) we have

Intervals of the domain where the  graph is above the x-axis

we know that

If the graph is above the x-axis, then the value of f(x) is positive

therefore

f(x) > 0

Part 2) we have

location on graph where input is zero  

Let

x ---> the independent variable or input value

f(x) ---> the dependent variable or output value

we know that

The y-intercept is the value of f(x) (output value) when the value of x (input value) is zero

therefore

y-intercept

Part 3) we have

location on graph where output is zero  

Let

x ---> the independent variable or input value

f(x) ---> the dependent variable or output value

we know that

The x-intercept is the value of x (input value) when the value of the function f(x) (output value) is zero

therefore

x-intercept

Part 4) we have

Intervals of the domain where the  graph is below the x-axis

we know that

If the graph is below the x-axis, then the value of f(x) is negative

therefore

f(x) < 0

c 1 hour

step-by-step explanation:

m = 2e

m = 2 so 2 = 2e

1 = e

3. (5x + 1) - (-10x + 6)

simplify the answer choice. combine like terms. remember to distribute the negative to the terms inside the second parenthesis. also remember that two negatives = one positive, and one of each sign = negative.

- (-10x + 6) = + 10x - 6

simplify. combine like terms:

5x + 1 + 10x - 6

5x + 10x + 1 - 6

15x - 5

∴ it is your answer

~