William asks…
how do you find an equation with slope?
I am new to this so could someone help me out
what is an equation of the line that passes through the point(3,-1)
and has the slope of 2 ?
admin answers:
The point–slope form of the equation for a line passing through the point (x1,y1) is found using: y-y1 = m(x-x1)
Assuming (3,-1) = (x1,y1) and m=2, and substituting the values we get: y-(-1)=2(x-3)
Simplified: y+1=2(x-3)
Helen asks…
How would you find an equation using point-slope from and slope-intercept form?
I NEED HELP. Okay, so Algebra exams are tomorrow. One question is:
Find the equation through (1,-7) and (4,-2) using point-slope form and slope intercept form.
I dont know how to do ANY of these. I know you find m which is 5/3, but what do you do from there on both? Please show step by step and make it as easy as possible. Thanks!!
admin answers:
Once you found the slope (m), which you've already correctly found for this problem, all you have to do is substitute into the general form to find: y = mx + b
This equation will hold for any point (x,y) on the line. Since we're looking at the line through those two points, we can use the x and y values from either point. Also notice that at this point, we have values for x, y, and m meaning that b is the only unknown, and we can easily find a value for it. For example,
-7 = (5/3)*(1) + b
or equally well, we could have done
-2 = (5/3)*(4) + b
Either way, if we solve for b, we'll get b = -26/3
So the equation of the line would be: y = (5/3)*x - (26/3)
Hope that helps. :)
Ken asks…
How do you rewrite an equation in standard form to slope intercept form?
I'm trying to catch up for a algebra test on Monday. One of the practice questions asks me to rewrite an equation that is in standard form to slope intercept form. I'm not exactly sure what it is asking me. I know the equation to find slope m = y2-y1/x2-x1, but I still do not know that the question is asking. The equation is an easy one, I just don't know what it wants me to do: 4x - 5y = 15
admin answers:
The slope intercept form is as follows:
y = mx + b
It is asking you to rearrange the following formula to adhere to the above line equation
4x - 5y = 15
add - 4x to both sides of the equation
4x - 4x - 5y = 15 - 4x
0 - 5y = 15 - 4x
divide both sides by - 5
y = 15/-5 - (4/-5)x
y = -3 + (4/5)x
The following is your slope y-intercept line equation:
y = (4/5)x - 3
which means the slope m = 4/5
and the y-intercept b = -3
Donna asks…
Point Slope. Find equation of line, parallel & perpendicular to the given line.?
ok so I know how to solve Point slope easily, but this one really bugs me to such an extent that I have to revise Point slope from scratch thus ruining my precious time.
Q. Find the equation of a line through P that is; a) parallel b) perpendicular to the given line.
i) P(-2,4) ; x=5
ii) P(-3,-2); y=3
My Solution:
i).
Method 1)
y-y1=m(x-x1)
y-4=m(5+2)
y-4=7m
(y-4)/7=m
Take m=m1
So for parallel, m2=(y-4)/7
For perpendicular m2= -7/(y-4)
y-y1=m(x-x1)
y-4=]-7/(y-4)](5+2)
(y-4)(y-4)= -49
y^2 -8y+16=-49
y^2 -8y=-65
Method 2)
Since only x is given, y must be 0, then; y-y1=m(x-x1)
0-4=m(5+2)
-4=7m
-4/7=m
Since m=m1
In parallel condition, m2=-4/7
In perpendicular m2= 7/4
y-y1=m(x-x1)
0-4=7/4(5+2)
4(-4)=7(7)
-16=49 = Screwed !
Method 3)
y=mx+b
0=5m +b
0=5+b
b=-5
Now,
y=mx+b
0=5m-5
5=5m
5/5=m
m=1
m=m1
Since parallel 1=m2
For Perpendicular, m2=-1
y-y1=m(x-x1)
y-4=-1(x+2)
y-4=-x-2
y= -x+2
Same 3 methods for part vi.But am so confused overall only for this question other wise I pawn at Slopes
admin answers:
Both of these are no solution. No real solution exists. When given "y = 3", you know it must be a horizontal line, because y is ALWAYS 3, according to your given equation. This can easily be proved, as the slope is zero. To determine slope of two given points, you put y2 - y1 over x2 - x1. If y is always the same number, than the top of your fraction which is defining your slope will equal zero, thus a horizontal line. However, a horizontal line which in which y is always 3, as determined by the given equation, cannot pass through the given point.
When x is given as a constant, it creates a vertical line. Do the slope formula and get zero in the denominator, thus, no slope. A vertical line that passes through the x axis at 5, and x is always always 5 cannot magically go through your given point.
Laura asks…
how to find slope and equation?
consider this function: y = 1/√x
(a) How do I find the slope of the tangent to the curve at the point where x = a?
(b) Also, how do I find the equation of the tangent line, f(x), at the point (4, 1/2)?
please explain, so that i can learn also..thank you!
admin answers:
(a)
The derivative is the same thing as the slope of a tangent line to the graph of a function. When we find the general form of the derivative then we are finding the general equation which describes the slope at any generic point x on the graph. So, first find the derivative of y = 1/√x:
y = 1/√x = x^(-½)
y' = -½ [x^(½ - 1)] = -½ [x^(-3/2)]
Since y' is equivalent to m, the slope of a tangent line to a function, then m = -½ [x^(-3/2)] for any x on the graph. So, if we let x = a, then the slope of the tangent line at that point is m = -½ [a^(-3/2)].
(b)
Since y' describes the slope of any given point on the graph of a function, then all we have to do to find the slope of the tangent line at any given point is plug the particular x value with which we are dealing into the equation for the derivative. We have done that generically above for this function. To find a particular numerical value for the tangent line, or slope, we fix the value of x numerically then calculate m. Since in this part of the problem we are given x = 4, then we plug 4 into the equation for the derivative to get this:
y' = m = -½ [x^(-3/2)]
y' = m = -½ [1/(4)^3/2] = -½ [1/(√4)³] = -½ [1/(2)³] = -½ (1/8) = -1/16.
Now that we have m, we need to find the y-intercept of the tangent line. Once we have that, we will have all the information we need to describe the equation of the tangent line. Use the slope-intercept form for the equation of a line, and plug in the given x and y values which lie on the graph of the function:
x = 4, y = ½
y = mx + b
½ = -1/16 (4) + b
½ = -4/16 + b
½ = -¼ + b
½ + ¼ = b
3/4 = b.
Now we know m and b, so we can write the equation of the tangent line at x = 4, for the graph of this function:
y = -1/16 x + 3/4.
Powered by Yahoo! Answers
No comments:
Post a Comment