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2 years ago

Find the slope of the line tangent to the function y=6-3x at the point (5,0)

Mathematics

1 answer:

choli [55]2 years ago

8 0

Answer:

-3

Step-by-step explanation:

To get the slope of the line tangent to the function y = 6 - 3x

First, we need to differentiate;

dy/dx = 0 -3x^{1-1}

dy/dx = 0 - 3x^0

dy/dx = 0 - 3

dy/dx = -3

Hence the slope of the line tangent to the function at the point is -3

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3 years ago

Graph this rational equation. Identify the points of discontinuity, holes, vertical asymptotes, x-intercepts, and horizontal asy

irakobra [83]

Step-by-step explanation:

We have given,

A rational function : f(x) = $\frac{x-2}{x-4}$

W need to find :

Point of discontinuity : - At x = 4, f(x) tends to reach infinity, So we get discontinuity point at x =4.

For no values of x, we get indetermined form (i.e $\frac{0}{0}$ ), Hence there is no holes

Vertical Asymptotes:

Plug y=f(x) = ∞ in f(x) to get vertical asymptote {We can us writing ∞ = $\frac{1}{0}$ }

i.e ∞ = $\frac{x-2}{x-4}$

or $\frac{1}{0}=\frac{x-2}{x-4}$

or x-4 =0

or x=4, Hence at x = 4, f(x) has a vertical asymptote

X -intercept :

Plug f(x)=0 , to get x intercept.

i.e 0 = $\frac{x-2}{x-4}$

or x - 2 =0

or x = 2

Hence at x=2, f(x) has an x intercept

Horizontal asymptote:

Plug x = ∞ in f(x) to get horizontal asymptote.

i.e f(x) = $\frac{x-2}{x-4}$ = $\frac{x(1-\frac{2}{x} )}{x(1-\frac{4}{x} )}$

or f(x) = $\frac{(1-\frac{2}{∞} )}{(1-\frac{4}{∞} )}$

or f(x) = 1 = y

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