1) Slope =-1, y-intercept = 0

Mathematics

1 answer:

Dima020 [189]2 years ago

6 0

Answer:

1. C 2.D 3.D 4.A

Step-by-step explanation:

The first one would be C because the slope is -1, which can also be -x and the y-intercept is 0, meaning you don't need to add or subtract anything.

The second one is D because the slope is -3, meaning it's -3x, and the y-intercept is positive, meaning you add 5.

The third one is D because the slope is 5/4, which is 5/4x and the y-intercept is positive 3, meaning you add 3.

And the fourth one is A because there is no slope, so y would just equal 3.

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For what values of x is f(x) = |x + 1| differentiable? I'm struggling my butt off for this course

pav-90 [236]

By definition of absolute value, you have

$f(x) = |x+1| = \begin{cases}x+1&\text{if }x+1\ge0 \\ -(x+1)&\text{if }x+1$

or more simply,

$f(x) = \begin{cases}x+1&\text{if }x\ge-1\\-x-1&\text{if }x$

On their own, each piece is differentiable over their respective domains, except at the point where they split off.

For x > -1, we have

(x + 1)' = 1

while for x < -1,

(-x - 1)' = -1

More concisely,

$f'(x) = \begin{cases}1&\text{if }x>-1\\-1&\text{if }x$

Note the strict inequalities in the definition of f '(x).

In order for f(x) to be differentiable at x = -1, the derivative f '(x) must be continuous at x = -1. But this is not the case, because the limits from either side of x = -1 for the derivative do not match:

$\displaystyle \lim_{x\to-1^-}f'(x) = \lim_{x\to-1}(-1) = -1$