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3 years ago

There is a line whose slope is 0 and whose y-intercept is 7. What is its equation in slope-intercept form?

Mathematics

1 answer:

daser333 [38]3 years ago

8 0

Step-by-step explanation:

To evaluate such, the following must be comprehended, on the behalf of linear data:

Slope: Rise/Run.

Y-intercept: The peculiar point in which the observed linear data intersects the y-axis.

X-intercept: The peculiar point in which the observed linear data intersects the x-axis.

Recall:

Slope-Intercept Form is acknowledged and defined as the integration of the intersection point, in relation or in proportion to the distance between two points within the linear data presented on the Cartesian Plane.

Slope-Intercept Form:

Y = mx + b

Y = The line.

M = Slope.

B = y-intercept.

The following may be equated, as stated:

- Slope = 0

Y = b

- Y-intercept = 7

Y = 7

Thus, on the Cartesian Plane is identified as a horizontal line positioned within quadrants I and II, intersection (0, 7).

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As distance must be positive, we reject the negative case, so x=8.

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(b) [-3,-2]

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(c) [-2,-1]

We plug in -2 for x and -1 for x

$f(-2) = (-2)^4 + 3(-2)^3 - 2(-2)^2 - 6(-2) - 1= -5$

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f(-2) is negative and f(-1) is positive. there is some value at x=c on the interval [-2,-1] where f(c)=0. so there exists atleast one zero on this interval.

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We plug in -1 for x and 0 for x

$f(-1) = (-1)^4 + 3(-1)^3 - 2(-1)^2 - 6(-1) - 1= 1$

$f(0) = (0)^4 + 3(0)^3 - 2(0)^2 - 6(0) - 1= -1$

f(-1) is positive and f(0) is negative. there is some value at x=c on the interval [-1,0] where f(c)=0. so there exists atleast one zero on this interval.

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We plug in 0 for x and 1 for x

$f(0) = (0)^4 + 3(0)^3 - 2(0)^2 - 6(0) - 1= -1$

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