When it is a line through the origin.
slopeintercept equations are used to find the slope and intercept (obviously lol) they are set up like this y=mx+b m is the slope and b is the y-intercept lets say you have an equation like... 2x + y = 5 (now minus 2x from both sides) 2x - 2x + y = 5 - 2x (simplify) y = 5 - 2x just use algebra to turn the standard form to slope intercept form
Just write the equation as: (x - 11)(x - 3) = 0 and convert it to any form you like.
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Solve it as you would normally. I could help if I had the equation. :) Fractions don't really change much; they just make the problem look more difficult.
Point-slope form is just another way to express a linear equation. It uses two (any two points that fall on the line) and the slope of the line (Therefore the name point-slope form).y2 - y1 = m(x2 - x1)...with m as the slope.
slopeintercept equations are used to find the slope and intercept (obviously lol) they are set up like this y=mx+b m is the slope and b is the y-intercept lets say you have an equation like... 2x + y = 5 (now minus 2x from both sides) 2x - 2x + y = 5 - 2x (simplify) y = 5 - 2x just use algebra to turn the standard form to slope intercept form
Before you can spot the slope, you absolutely must massage the equation so thatit's in the right form. The right form isy = Ax + B . . . and ' A ' and ' B ' are just numbers.When your equation is in that form, then ' A ' is the slope of the graph, and ' B ' is the y-intercept.
The slope-intercept form of an equation is: y = mx + b Just copy down this equation, then replace "m" with the slope, and "b" with the y-intercept.
Just write the equation as: (x - 11)(x - 3) = 0 and convert it to any form you like.
No, it is not even an equation - just a single term!No, it is not even an equation - just a single term!No, it is not even an equation - just a single term!No, it is not even an equation - just a single term!
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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.
Solve it as you would normally. I could help if I had the equation. :) Fractions don't really change much; they just make the problem look more difficult.
Point-slope form is just another way to express a linear equation. It uses two (any two points that fall on the line) and the slope of the line (Therefore the name point-slope form).y2 - y1 = m(x2 - x1)...with m as the slope.
(3, 18), (3, 18) is just one point: it does not define a line.
B is just a constant variable. In the standard form for the equation of a line y= ax +b. When x = 0, this is where we cross the y axis and the equation evaluates to b.
a quadratic equation must be in this form ax^2+bx+c=0 (can either be + or -) an exponential just means that the function grows at an exponential rate f(x)=x^2 or x^3