Question about Texas Instruments TI-84 Plus Calculator

I couldn't find how to find the slope in an equation from any other sites.

Ad

If your equation can be written in the form of a function (y=f(x)),

the slope of the function at a point is the derivative of the function at the point. For any function that is not a constant or linear, you need calculus to find the slope.

If the function is of the form y= ax +b, the slope is the value a.

Posted on Feb 11, 2016

Ad

Hi,

a 6ya expert can help you resolve that issue over the phone in a minute or two.

best thing about this new service is that you are never placed on hold and get to talk to real repairmen in the US.

the service is completely free and covers almost anything you can think of (from cars to computers, handyman, and even drones).

click here to download the app (for users in the US for now) and get all the help you need.

goodluck!

Posted on Jan 02, 2017

Ad

The easiest way to solve and graph and equation is to put the equation into the slope intercept form y = mx + b, where m is the slope and b in the y-intercept.

To do this, we subtract 4x from both sides and get y = -4x + 0.

From this we know m = -4 (slope) and the y-intercept is 0.

I always start with the y-intercept and put a point there. Thus, we have a point at (0,0). Using this as a starting point, we now use the slope of -4 to get future points. Since it is negative, we go one unit to the left and up four units. So we have the point (-1,4). Using a ruler, we connect these points and continue on both sides to produce the line.

There is also a great free online program/app called Desmos that you can use to check your work. Type in the equation of the line and it will graph it for you.

Good luck,

Paul

To do this, we subtract 4x from both sides and get y = -4x + 0.

From this we know m = -4 (slope) and the y-intercept is 0.

I always start with the y-intercept and put a point there. Thus, we have a point at (0,0). Using this as a starting point, we now use the slope of -4 to get future points. Since it is negative, we go one unit to the left and up four units. So we have the point (-1,4). Using a ruler, we connect these points and continue on both sides to produce the line.

There is also a great free online program/app called Desmos that you can use to check your work. Type in the equation of the line and it will graph it for you.

Good luck,

Paul

Oct 27, 2016 | Office Equipment & Supplies

The equation of any straight line, called a linear equation, can be written as: **y** = mx + b, where m is the slope of the line and b is the **y**-**intercept**. The **y**-**intercept** of this line is the value of **y** at the point where the line crosses the **y** axis.

Graphing Equations and Inequalities Slope and intercept First Glance

Graphing Equations and Inequalities Slope and intercept First Glance

Feb 05, 2016 | Apple iPod touch

http://www.webmath.com/equline3.html

May 20, 2015 | Office Equipment & Supplies

A great tutorial: https://www.khanacademy.org/math/cc-eighth-grade-math/cc-8th-linear-equations-functions/8th-slope-intercept-form/v/linear-equations-in-slope-intercept-form

Mar 17, 2015 | math.com Computers & Internet

The "slope intercept form of the equation of a (the) line" is y=mx+b, where m is the slope of the line and b is the y-intercept.

We are given the slope of 1/2, so m= 1/2.

We can now write y=1/2 x + b.

Since the point (-2,-3) is on the line, we can substitute it in and solve for b. We put the -2 in for x and -3 in for y.

-3 = 1/2(-2) +b

-3 = -1 + b

-3 + 1 = -1 + b +1

-2 =b

Thus, the equation of the line is y= 1/2 x -2

To check if we did this correctly, plug in the point (-2, -3) to see if it works.

Left Side Right Side

-3 = 1/2 (-2) -2

= -1-2

= -3

We are given the slope of 1/2, so m= 1/2.

We can now write y=1/2 x + b.

Since the point (-2,-3) is on the line, we can substitute it in and solve for b. We put the -2 in for x and -3 in for y.

-3 = 1/2(-2) +b

-3 = -1 + b

-3 + 1 = -1 + b +1

-2 =b

Thus, the equation of the line is y= 1/2 x -2

To check if we did this correctly, plug in the point (-2, -3) to see if it works.

Left Side Right Side

-3 = 1/2 (-2) -2

= -1-2

= -3

Feb 24, 2015 | Office Equipment & Supplies

That's a little difficult since there is no y in the equation.

Nov 19, 2013 | Computers & Internet

That is an equation describing a straight line. The "slope-intercept" form of a line is

y = mx + b

where m is the slope (change in y-value / change in x-value)

and b is the y-intercept (the point where the line crosses the y-axis when x=0)

Positive slope means the line is rising and negative slope means it's falling.

You can rewrite the original equation 2x - 4y -9 = 0 in slope-intercept form:

y = (1/2)x - (9/4)

So you know the slope is positive 1/2 (line rises 1 y-unit for each 2 x-unit change) and crosses the y-axis at -9/4. With this information you can graph the line.

y = mx + b

where m is the slope (change in y-value / change in x-value)

and b is the y-intercept (the point where the line crosses the y-axis when x=0)

Positive slope means the line is rising and negative slope means it's falling.

You can rewrite the original equation 2x - 4y -9 = 0 in slope-intercept form:

y = (1/2)x - (9/4)

So you know the slope is positive 1/2 (line rises 1 y-unit for each 2 x-unit change) and crosses the y-axis at -9/4. With this information you can graph the line.

Jul 12, 2011 | Sewing Machines

The site seems to eat the plus signs I enter, so I will use PLUS to symbolize addition.

To find the equation of the straight line (y = a*x PLUS b) that passes through two points P1(x1,y1) and P(x2,y2) , you need to use

1. the coordinates of the points to calculate the slope a (gradient) as a=(y2-y1)/(x2-x1)

2. Replace the calculated value of a in the equation and write that one of the points ( P1(x1,y1) for example) satisfies the equation. In other words y1=a*x1 PLUS b.

Here y1 and x1 are known values, a has been calculated, and only b is still unknown. You can now use the equation y1=a*x1 PLUS b to calculate b as

b=(y1-a*x1)

Example: Equation of the line through (1,5) and (3,6)

Calculate the slope (gradient) of the line as a=(y2-y1)/(x2-x1) where y2=6, y1=5, x2=3, and x1=1. You should get a=(6-5)/(3-1)=1/2

The equation is y=(1/2)x PLUS b, where b is not known yet.

To find b, substitute the coordinates of one of the points in the equation. Let us do it for (3,6).

The point (3,6) lies on the line, so 6=(1/2)*3 PLUS b.

Solve for b: 6 MINUS 3/2=b, or b=9/2=4.5

Equation is thus y=(x/2) PLUS 9/2 =(x PLUS 9)/2

I trust you can substitute you own values for (x1,y1, x2,y2) to duplicate the calculations above.

To find the equation of the straight line (y = a*x PLUS b) that passes through two points P1(x1,y1) and P(x2,y2) , you need to use

1. the coordinates of the points to calculate the slope a (gradient) as a=(y2-y1)/(x2-x1)

2. Replace the calculated value of a in the equation and write that one of the points ( P1(x1,y1) for example) satisfies the equation. In other words y1=a*x1 PLUS b.

Here y1 and x1 are known values, a has been calculated, and only b is still unknown. You can now use the equation y1=a*x1 PLUS b to calculate b as

b=(y1-a*x1)

Example: Equation of the line through (1,5) and (3,6)

Calculate the slope (gradient) of the line as a=(y2-y1)/(x2-x1) where y2=6, y1=5, x2=3, and x1=1. You should get a=(6-5)/(3-1)=1/2

The equation is y=(1/2)x PLUS b, where b is not known yet.

To find b, substitute the coordinates of one of the points in the equation. Let us do it for (3,6).

The point (3,6) lies on the line, so 6=(1/2)*3 PLUS b.

Solve for b: 6 MINUS 3/2=b, or b=9/2=4.5

Equation is thus y=(x/2) PLUS 9/2 =(x PLUS 9)/2

I trust you can substitute you own values for (x1,y1, x2,y2) to duplicate the calculations above.

Jan 27, 2011 | Texas Instruments TI-84 Plus Calculator

Sorry we dont do homework its cheating

Sep 24, 2009 | Riverdeep Mighty Math Cosmic Geometry...

assuming the question is what is the circle equation?

and if (-2,2) is the center of the circle

the equation should look like this: (x+2)^2+(Y-2)^2=R^2

And now only R is needed.

given 2x-5y+4=0 equation of line perpendicular

we can rearange the equation to be y=(2x+4)/5

from that we can see that the slope of the line is 2/5

And from the fact of perpendicular line we can say that the slope

of the radius line is -2/5.

The motivation now is to calculate the distance between the center of the circle to the cross point of the radius with the line perpendicular

For that we would calculate the radius line equation and compare it to the equation of line perpendicular

As mentioned earlier the slope of the radious line is -2/5.

So the equation is y=-2/5x+b and b can be calculated by using the center of the circle coordinates

2= - (2/5)*(-2)+b ------> b=2-4/5=1.2

radius equation is y=-(2/5)x+1.2

Now the cross point is calculated by comparing the equations:

-(2/5)x+1.2=(2x+4)/5 --> -2x+6=2x+4 --> 4x=2 --> x=1/2 --> y=1

So the cross point is (1/2,1).

The distance between the points is calculated by the following

Formula:

R=SQR(((1/2)-(-2))^2+(2-1)^2)=SQR(2.5^2+1^2)=SQR(6.25+1)=

SQR(7.25)

Therefore the circle eq is (x+2)^2+(Y-2)^2=7.25

and if (-2,2) is the center of the circle

the equation should look like this: (x+2)^2+(Y-2)^2=R^2

And now only R is needed.

given 2x-5y+4=0 equation of line perpendicular

we can rearange the equation to be y=(2x+4)/5

from that we can see that the slope of the line is 2/5

And from the fact of perpendicular line we can say that the slope

of the radius line is -2/5.

The motivation now is to calculate the distance between the center of the circle to the cross point of the radius with the line perpendicular

For that we would calculate the radius line equation and compare it to the equation of line perpendicular

As mentioned earlier the slope of the radious line is -2/5.

So the equation is y=-2/5x+b and b can be calculated by using the center of the circle coordinates

2= - (2/5)*(-2)+b ------> b=2-4/5=1.2

radius equation is y=-(2/5)x+1.2

Now the cross point is calculated by comparing the equations:

-(2/5)x+1.2=(2x+4)/5 --> -2x+6=2x+4 --> 4x=2 --> x=1/2 --> y=1

So the cross point is (1/2,1).

The distance between the points is calculated by the following

Formula:

R=SQR(((1/2)-(-2))^2+(2-1)^2)=SQR(2.5^2+1^2)=SQR(6.25+1)=

SQR(7.25)

Therefore the circle eq is (x+2)^2+(Y-2)^2=7.25

Oct 26, 2008 | Casio FX-115ES Scientific Calculator

66 people viewed this question

Usually answered in minutes!

×