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3 years ago

What is the equation of the line that is perpendicular to the given line and has an x-intercept of -3?

Mathematics

1 answer:

Anvisha [2.4K]3 years ago

3 0

Answer:

$y=-\frac{3}{2}x-\frac{9}{2}$

Step-by-step explanation:

Let the equation of the perpendicular line is,

y = mx + b

where m = slope of the line

b = y-intercept

From the graph, slope of the line passing through (0, -1) and (3, 1),

m' = $\frac{y_2-y_1}{x_2-x_1}$

m' = $\frac{1+1}{3-0}$

m' = $\frac{2}{3}$

To get the slope (m) of this line we will use the property of perpendicular lines,

m × m' = (-1)

m × $\frac{2}{3}$ = -1

m = $-\frac{3}{2}$

Equation of the perpendicular line will be,

$y=-\frac{3}{2}x+b$

x-intercept of the line is (-3) therefore, point on the line is (-3, 0)

0 = $-\frac{3}{2}(-3)+b$

b = $-\frac{9}{2}=-4.5$

Equation of the line will be,

$y=-\frac{3}{2}x-\frac{9}{2}$

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3 years ago

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Part A

The notation $\lim_{x \to 2^{+}}f(x)$ means that we're approaching x = 2 from the right hand side (aka positive side). This is known as a right hand limit.

So we could start at say x = 2.5 and get closer to 2 by getting to x = 2.4 then to x = 2.3 then 2.2, 2.1, 2.01, 2.001, etc

We don't actually arrive at x = 2 itself. We simply move closer and closer.

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Therefore f(x) = (x/2) + 1

Plug in x = 2 to find that...

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Then for the left hand limit $\lim_{x \to 2^{-}}f(x)$ , we'll involve x < 2 and we go for the first piece. So,

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Therefore, $\lim_{x \to 2^{-}}f(x) = 1$

===============================================================

Part B

Because $\lim_{x \to 2^{+}}f(x) \ne \lim_{x \to 2^{-}}f(x)$ this means that the limit $\lim_{x \to 2}f(x)$ does not exist.

If you are a visual learner, check out the graph below of the piecewise function. Notice the gap or disconnect at x = 2. This can be thought of as two roads that are disconnected. There's no way for a car to go from one road to the other. Because of this disconnect, the limit doesn't exist at x = 2.

===============================================================

Part C

You'll follow the same type of steps shown in part A.

However, keep in mind that x = 4 is above x = 2, so we'll deal with x > 2 only.

So you'd only involve the second piece f(x) = (x/2) + 1

You should find that f(4) = 3, and that both left and right hand limits equal this value. The left and right hand limits approach the same y value. The limit does exist here. There are no gaps to worry about when x = 4.

===============================================================

Part D

As mentioned earlier, since $\lim_{x \to 4^{+}}f(x) = \lim_{x \to 4^{-}}f(x) = 3$ , this means the limit $\lim_{x \to 4}f(x)$ does exist and it's equal to 3.