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A normal slope-intercept form equation would read: y = mx + b The slope of an equation is also known as 'm'. The y-intercept would count as 'b'. So in a random equation such as: y= 5x + 6 '5' would be the slope and '6' would be the y-intercept.

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y=-4

You have to differentiate the equation. The dy/dx is the slope.

first of all, i dont see how slope and yintercept has anything to do with shopping. usually in an equation, y= mx+b, m stands for slope, and b is the y-intercept. x and y are just x and y. If the equation is mixed up, and is not in y=mx+b format, you need to solve to get y on one side, the slope, x, and whatever the presented constant is (which is the y-intercept) on the other side. hope this helps.

An equation of a line requires two parameters. The slope, by itself, is not enough.

the slope formula is y=mx+b slope-intercept form of an equation of a line. where m=slope and b=the y-intercept

Use point-slope formula

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Slope or gradient = (y2-y1)/(x2-x1)

the slope is 0

As for example in the straight line equation of y=3x+5 the slope is 3 and the y intercept is 5

y=mx+b whatever multiplies x in this form is the slope

The slope is the gradient which is y2-y1/x2-x1 To find the y int you set the equation of the line equal to zero i think :/

the slope is the 'm' in y=mx+b so even if the points aren't given, if there is an equation, then you can find the slope. for example, if you have an equation like this: y=2x+5 the slope is 2 and the y-intercept is 5.

This is the first fundemental theorem of Calculus. The slope of a line is very important in your first calculus course. The slope tells you the rate of change. This means how much is the object change in height compared to its change in length. The slope of a line in Calculus is used as the first derivative. If you can take the slope of a line at one particular point you will find the answer to the derivative at this point. Remember this. You first equation on your graph is called your position equation. If you take the derivative of this equation it is called the velocity equation. The velocity equation is how much the position equation is sloping at each point. If you take the derivative of the velocity equation you will get the acceleration equation. The accerelation equation is how much the velocity is sloping at each point. You can take the derivative of the acceleration equation and this will give you the jerk equation. The jerk equation is not used in many applications and I have never used this equation in any of my 4 calculus classes.

we have to change the given equation to y=mx+b.here m is the slope

The slope of a straight line equation is: y2-y1/x2-x1

If it is the equation for a line, then it can be rearranged into the format y = mx + b, where m is the slope of the line, and b is the point where the line intercepts the y-axis.If it is not for a straight line, then the slope is changing with x, and the derivative of the function would find the slope at a particular x.

3

We know that the line passes through points (2, 2) and (0, 10) (since the y-intercept is 10).Using these two points, we can find the slope of the line,m = (10 - 2)/(0 - 2) = 8/-2 = 4/-1 = -4.Now by using the slope, m = -4, and the y-intercept, 10, we can write the equation of the line in the slope-intercept form, y = mx + b which isy = -4x + 10.

Use: (y2 -y1)/(x2 -x1) to find the slope. Use: y -y1 = m(x -x1) to find the slope intercept equation whereas m is the slope.

You need to find the slope in the equation.

The m is the slope in the y=mx+b or the number before the x.

The equation for the slope of a line is y=mx+b

Here is how to solve it. First, find the slope of the given line. To do this, solve the equation for "y". That will convert the equation to the slope-intercept form. From there, you can immediately read off the slope. Since parallel lines have the same slope, the line you are looking for will have the same slope. Now you need to use the point-slope form of the equation, with the given point, and the slope you just calculated. Finally, solve this equation for "y" to bring it into the requested slope-intercept form.