Question

Two linear functions are displayed. Function One is represented by the graph:

Answer 1

Answer:

Option A

Step-by-step explanation:

Function One is represented by the graph. We can see that the slope of the First Function is positive because the formula of slope is

$Slope=\frac{Change\ in\ y}{Change\ in\ x}$

And we can see that in the graph the change in y and change in x are both positive as the function is increasing because if we see the graph closely,

When x = 1 the value of y = 2 and when x = 2 the value of y = 4 and when x = 3 the value of y = 6 so the values are increasing of x and of y hence the slope is positive and the value of slope is 2.

Now the Second Function which is y = -4x here the slope is negative because the slope-intercept form of the equation is .

$y=mx+b$

and the second function is

$y=-4x$

if we compare both these equations we get the value of m = -4 where m is also called the slope. So the slope of the second function is negative.

So

Option B is incorrect because the slope of the first function is positive

Option C is incorrect because the slope of the second function is negative

Option D is incorrect because the slope of the first function is 2 and the slope of the second function is -4

That leaves us with Option A which is correct, where we can see that the slope of the first function is positive and the slope of the second function is negative

Answer 2

Answer:

A) The slope of function one is positive and the slope of function two is negative.

General Formulas and Concepts:

Algebra I

Slope-Intercept Form: y = mx + b

• m - slope
• b - y-intercept

Step-by-step explanation:

We see that the 1st graph has a positive slope; as x increases, the y-value also increases. Therefore, the 1st graph must have a positive slope.

We are given the equation y = -4x as our second line. Using our y = mx + b form, we see that our slope m = -4. Therefore, the 2nd graph has a negative slope.

Our answer choice then would be A, as it is the only one that matches the descriptions.