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lions [1.4K]
3 years ago
5

What is the equation, in point-slope form, of the line that has a slope of 6 and passes through the point (–1, –8)? a. y+8 = 6 (

x+1 )
Mathematics
2 answers:
Stolb23 [73]3 years ago
8 0

Answer:

y + 8 = 6(x + 1)

Step-by-step explanation:

The equation of a line in point- slope form is

y - b = m(x - a)

where m is the slope and (a, b) a point on the line

here m = 6 and (a, b) = (- 1, - 8), hence

y - (- 8) = 6(x - (- 1)), that is

y + 8 = 6(x + 1)

rodikova [14]3 years ago
4 0

<u>ANSWER</u>

y + 8=6(x + 1)

<u>EXPLANATION</u>

The point-slope form of an equation is given by the formula:

y-y_1=m(x-x_1)

Where

(x_1,y_1)

is a point on this line and

m

is the slope of the line.

From the question, the line has slope 6 and passes through (-1,-8).

This means that

x_1=-1

y_1=-8

and

m = 6

We substitute these values into the point-slope formula to get:

y- - 8=6(x- - 1)

We simplify to get:

y + 8=6(x + 1)

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Consider the equation below. (If an answer does not exist, enter DNE.) f(x) = x3 − 6x2 − 15x + 4 (a) Find the interval on which
kozerog [31]

Answer:

a) The function, f(x) is increasing at the intervals (x < -1.45) and (x > 3.45)

Written in interval form

(-∞, -1.45) and (3.45, ∞)

- The function, f(x) is decreasing at the interval (-1.45 < x < 3.45)

(-1.45, 3.45)

b) Local minimum value of f(x) = -78.1, occurring at x = 3.45

Local maximum value of f(x) = 10.1, occurring at x = -1.45

c) Inflection point = (x, y) = (1, -16)

Interval where the function is concave up

= (x > 1), written in interval form, (1, ∞)

Interval where the function is concave down

= (x < 1), written in interval form, (-∞, 1)

Step-by-step explanation:

f(x) = x³ - 6x² - 15x + 4

a) Find the interval on which f is increasing.

A function is said to be increasing in any interval where f'(x) > 0

f(x) = x³ - 6x² - 15x + 4

f'(x) = 3x² - 6x - 15

the function is increasing at the points where

f'(x) = 3x² - 6x - 15 > 0

x² - 2x - 5 > 0

(x - 3.45)(x + 1.45) > 0

we then do the inequality check to see which intervals where f'(x) is greater than 0

Function | x < -1.45 | -1.45 < x < 3.45 | x > 3.45

(x - 3.45) | negative | negative | positive

(x + 1.45) | negative | positive | positive

(x - 3.45)(x + 1.45) | +ve | -ve | +ve

So, the function (x - 3.45)(x + 1.45) is positive (+ve) at the intervals (x < -1.45) and (x > 3.45).

Hence, the function, f(x) is increasing at the intervals (x < -1.45) and (x > 3.45)

Find the interval on which f is decreasing.

At the interval where f(x) is decreasing, f'(x) < 0

from above,

f'(x) = 3x² - 6x - 15

the function is decreasing at the points where

f'(x) = 3x² - 6x - 15 < 0

x² - 2x - 5 < 0

(x - 3.45)(x + 1.45) < 0

With the similar inequality check for where f'(x) is less than 0

Function | x < -1.45 | -1.45 < x < 3.45 | x > 3.45

(x - 3.45) | negative | negative | positive

(x + 1.45) | negative | positive | positive

(x - 3.45)(x + 1.45) | +ve | -ve | +ve

Hence, the function, f(x) is decreasing at the intervals (-1.45 < x < 3.45)

b) Find the local minimum and maximum values of f.

For the local maximum and minimum points,

f'(x) = 0

but f"(x) < 0 for a local maximum

And f"(x) > 0 for a local minimum

From (a) above

f'(x) = 3x² - 6x - 15

f'(x) = 3x² - 6x - 15 = 0

(x - 3.45)(x + 1.45) = 0

x = 3.45 or x = -1.45

To now investigate the points that corresponds to a minimum and a maximum point, we need f"(x)

f"(x) = 6x - 6

At x = -1.45,

f"(x) = (6×-1.45) - 6 = -14.7 < 0

Hence, x = -1.45 corresponds to a maximum point

At x = 3.45

f"(x) = (6×3.45) - 6 = 14.7 > 0

Hence, x = 3.45 corresponds to a minimum point.

So, at minimum point, x = 3.45

f(x) = x³ - 6x² - 15x + 4

f(3.45) = 3.45³ - 6(3.45²) - 15(3.45) + 4

= -78.101375 = -78.1

At maximum point, x = -1.45

f(x) = x³ - 6x² - 15x + 4

f(-1.45) = (-1.45)³ - 6(-1.45)² - 15(-1.45) + 4

= 10.086375 = 10.1

c) Find the inflection point.

The inflection point is the point where the curve changes from concave up to concave down and vice versa.

This occurs at the point f"(x) = 0

f(x) = x³ - 6x² - 15x + 4

f'(x) = 3x² - 6x - 15

f"(x) = 6x - 6

At inflection point, f"(x) = 0

f"(x) = 6x - 6 = 0

6x = 6

x = 1

At this point where x = 1, f(x) will be

f(x) = x³ - 6x² - 15x + 4

f(1) = 1³ - 6(1²) - 15(1) + 4 = -16

Hence, the inflection point is at (x, y) = (1, -16)

- Find the interval on which f is concave up.

The curve is said to be concave up when on a given interval, the graph of the function always lies above its tangent lines on that interval. In other words, if you draw a tangent line at any given point, then the graph seems to curve upwards, away from the line.

At the interval where the curve is concave up, f"(x) > 0

f"(x) = 6x - 6 > 0

6x > 6

x > 1

- Find the interval on which f is concave down.

A curve/function is said to be concave down on an interval if, on that interval, the graph of the function always lies below its tangent lines on that interval. That is the graph seems to curve downwards, away from its tangent line at any given point.

At the interval where the curve is concave down, f"(x) < 0

f"(x) = 6x - 6 < 0

6x < 6

x < 1

Hope this Helps!!!

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3 years ago
Find B if a= 8M, B=69°, and C=48°​
dmitriy555 [2]

Answer:

Step-by-step explanation:

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3 years ago
(-3,-14) and (0-9)<br> Pls help!!!
andreyandreev [35.5K]

Answer:

<em>y</em> = (5/3)<em>x</em> - 9

Step-by-step explanation:

First, find the slope (rise over run):

m = (-9 - (-14)) / (0 - (-3)) = (-9 + 14) / (0 + 3) = 5/3

Use the point-slope form equation with the given points (-3, -14):

y + 14 = (5/3)(x + 3)

or optionally in slope intercept form:

<em>y</em> = (5/3)<em>x</em> - 9

3 0
2 years ago
Which expressions are equivalent to 6g-18h
sesenic [268]

Answer:

6(g-3h)

Step-by-step explanation:

6/6=1

18/6=3

6(g-3h)

5 0
3 years ago
Read 2 more answers
Evelyn pets the box with 1 inch cubes with represents does not show how evelyn can’t find the volume of the box
erastovalidia [21]

Question Correction

Evelyn packed this box with 1 inch cubes. Which expression does not show how Evelyn can find the volume of the box?

  • (A)6+6+6+6+6+6
  • (B) 2 X 3 X 6
  • (C) 2 + 3 + 6
  • (D) 6 X 6

Answer:  

(C) 2 + 3 + 6

Step-By-Step Explanation

In the diagram,  

  • Height = 6 Units
  • Length =2 Units
  • Width =3 Units

Volume  = Height X Length X Width

= 6 X 2 X 3

=36 cubic units

Consider the options:

(A)6+6+6+6+6+6 =36 cubic units

(B) 2 X 3 X 6 =36 cubic units

(C) 2 + 3 + 6 = 11 cubic units

(D) 6 X 6 =36 cubic units

Out of the option, that which is not equivalent to 36 cubic units is:

(C) 2+3+6

Therefore, it does not show how Evelyn can find the volume of the box.

6 0
2 years ago
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