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2 years ago

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On a coordinate plane, a line goes through (0, negative 1) and (3, 1). a point is at (negative 3, 0). what is the equation of th

e line that is parallel to the given line and has an x-intercept of â€"3? y = two-thirdsx 3 y = two-thirdsx 2 y = negative three-halvesx 3 y = â€"three-halvesx 2

1 answer:

belka [17]2 years ago

8 0

The equation of the line that is parallel to the given line and has an x-intercept of 3 is y = 2/3x - 2

<h3>What is the equation of the line that is parallel to the given line and has an x-intercept of 3?</h3>

The given parameters are

Point =(0, -1) and (3, 1)

Calculate the slope using

m = (y₂ - y₁)/(x₂ - x₁)

Where m represents the slope

So, we have:

m = (1 -- 1)/(3 -0)

Evaluate

m = 2/3

Parallel lines have equal slopes

So, the slope of the other line is

m = 2/3

Linear equations are represented as:

y = mx + c

Where m represents the slope and c represents the y-intercept

So, we have

y = 2/3x + c

An x-intercept of 3 means

0 = 2/3 * 3 + c

So, we have

c = -2

Substitute c = -2 in y = 2/3x + c

y = 2/3x - 2

Hence, the equation of the line that is parallel to the given line and has an x-intercept of 3 is y = 2/3x - 2

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------------Formal Definition of Derivative----------------

The following limit will give us the derivative of the function $f(x)=2x^2-7x+6$ at $x=4$ (the slope of the tangent line at $x=4$ ):

$\lim_{x \rightarrow 4}\frac{f(x)-f(4)}{x-4}$

$\lim_{x \rightarrow 4}\frac{2x^2-7x+6-10}{x-4}$ We are given f(4)=10.

$\lim_{x \rightarrow 4}\frac{2x^2-7x-4}{x-4}$

Let's see if we can factor the top so we can cancel a pair of common factors from top and bottom to get rid of the x-4 on bottom:

$2x^2-7x-4=(x-4)(2x+1)$

Let's check this with FOIL:

First: $x(2x)=2x^2$

Outer: $x(1)=x$

Inner: $(-4)(2x)=-8x$

Last: $-4(1)=-4$

---------------------------------Add!

$2x^2-7x-4$

So the numerator and the denominator do contain a common factor.

This means we have this so far in the simplifying of the above limit:

$\lim_{x \rightarrow 4}\frac{2x^2-7x-4}{x-4}$

$\lim_{x \rightarrow 4}\frac{(x-4)(2x+1)}{x-4}$

$\lim_{x \rightarrow 4}(2x+1)$

Now we get to replace x with 4 since we have no division by 0 to worry about:

2(4)+1=8+1=9.

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