What is the slope of a line that is perpendicular to the line y = 1?

2 answers:

8 0

When u have an equation such as y = 1......this means that u have a horizontal line which has a slope of 0.

* now if u had an equation such as x = 1...this would be a vertical line with an undefined slope....just so u know.

Paha777 [63]3 years ago

4 0

Answer:

Undefined ( ∞ )

Step-by-step explanation:

The given line is y=1

It is horizontal line on coordinate plane.

Because y=k means y become constant and x varies.

If we draw <em>perpendicular</em> to horizontal line then it would be vertical line.

The equation of vertical line, x=h

The slope of horizontal line is 0

The slope of vertical line is undefined.

Because vertical line makes angle with x-axis 90°

Slope = tan 90°

∵ tan 90° is undefined.

Hence, The slope of perpendicular line of y=1 is undefined.

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I need help with part b. I feel like there’s a catch, I want to do the first derivative test, however, I feel like there is a be

Sladkaya [172]

Answer:

The fifth degree Taylor polynomial of g(x) is increasing around x=-1

Step-by-step explanation:

Yes, you can do the derivative of the fifth degree Taylor polynomial, but notice that its derivative evaluated at x =-1 will give zero for all its terms except for the one of first order, so the calculation becomes simple:

$P_5(x)=g(-1)+g'(-1)\,(x+1)+g"(-1)\, \frac{(x+1)^2}{2!} +g^{(3)}(-1)\, \frac{(x+1)^3}{3!} + g^{(4)}(-1)\, \frac{(x+1)^4}{4!} +g^{(5)}(-1)\, \frac{(x+1)^5}{5!}$

and when you do its derivative:

1) the constant term renders zero,

2) the following term (term of order 1, the linear term) renders: $g'(-1)\,(1)$ since the derivative of (x+1) is one,

3) all other terms will keep at least one factor (x+1) in their derivative, and this evaluated at x = -1 will render zero

Therefore, the only term that would give you something different from zero once evaluated at x = -1 is the derivative of that linear term. and that only non-zero term is: $g'(-1)= 7$ as per the information given. Therefore, the function has derivative larger than zero, then it is increasing in the vicinity of x = -1