Help With Math!!!?

Question

line L is tangent to the graph of y=x-(x^2/500) at point q. one point on line L is (0,20).
a) find the x-coordinate of point q
b)write an equation for line L

line L is tangent to the graph of y=x-(x^2/500) at point q. one point on line L is (0,20), but this is not the tangent point (point q), it is a random point on the line to help you solve the problem, i hope that clears things up a little bit.
a) find the x-coordinate of point q
b)write an equation for line L

Answer 1

a) ± 100
b) y = (3/5)x + 20 or y = (7/5)x + 20

let q be (a,b), so we know that b = a - (a^2/500)

if y=x-(x^2/500) then y' = 1 - x/250
at point q, the slope is then 1 - a/250

Since (0,20) is on the line, the slope of the line is also (b-20)/(a-0) = (b-20)/a

so 1 - a/250 = (b-20)/a
or a - a^2/250 = b - 20
and b = a - (a^2/500) from above.

a - a^2/250 = a - (a^2/500) - 20
a^2/500 = 20
a^2 = 10000
a = ±100

line 1:
point q (100, 80)
point L (0,20)
y = mx + b ; m = 60/100 = 3/5, b = 20
y = (3/5)x + 20 or y = 0.6x + 20

line 2:
point q (-100,-120)
point L (0,20)
y = mx + b ; m = 140/100 = 7/5, b = 20
y = (7/5)x + 20 or y = 1.4x + 20

******************* ******************* *******************
I DONT UNDERSTAND YOUR ADDITIONAL DETAILS. Didn't you read my answer? It is correct.
Wil T almost has it right. He got one of the x-coordinates, but neglected the answer -100. (There are two square roots of 10000: 100 and -100). His line equation is incorrect however, as he did not solve for the y-coordinate of q correctly. The point q is (100, 80) not (100,40).
You can also use my numbers to go back and double check.
We established that the slope of the tangent line is 1 - a/250. If a = 100, the slope is 1 - 100/250 = 3/5, which is indeed the slope of the line y=(3/5)x + 20. If a = -100, the slope is
1 - (-100)/250 = 7/5, which is the slope of the line
y = (7/5)x + 20.

******************* ******************* *******************
▼▼▼▼▼▼▼▼▼▼▼▼▼▼▼▼▼▼▼▼▼▼▼▼
SEE:
http://mudandmuck.com/str2/calc1.jpg
▲▲▲▲▲▲▲▲▲▲▲▲▲▲▲▲▲▲▲▲▲▲▲▲

Answer 2

First off, I'd get a second opinion on this...

Secondly, what level math is this? This is hard!!
Anyway here goes…
Let's call point q (a,b)
- you need to know that the gradient of the tangent and the gradient of the graph will be the same at the tangent point.

- You also need to know that the gradient of a line is (y2-y1)/(x2-x1) where (x1,y1) and (x2,y2) are points on the line.

You also need to know that dy/dx will give you the gradient function for the graph at any given point.

You also need to know that if y=x then dy/dx=1 and if y=x^n then dy/dx=nx^(n-1) i.e you need to be able to differentiate.

So, being that you know all that let's do this.

y= x- (x^2/500)
dy/dx = 1-(x/250) at the tangent point, x=a you with me?
So gradient of graph at tangent point is 1-(a/250)

Gradient of tangent- b-20/a-0…given that (0,20) is a point on the line.

Therefore we can say: 1-(a/250) =b-20/a-0
Which boils down to b=a-(a^2/250)+20

Now at the point where the tangent touches the graph is (a,b) so from the equation of the graph  y= x- (x^2/500) we can say b= a-(a^2/500)

So substituting for b above gives

a-(a^2/500)=a-(a^2/250)+20

This works out to be a=100  x-coordinate of the point q

Now if b= a-(a^2/250)+20 and a=100 then
b= 40
Therefore q is (100,40)

Gradient of tangent 40-20/100-0 = 1/5

Equation of tangent is of the form y=mx+c where m is the gradient and c is where the line crosses the y-axis (we know this is at 20)

So equation is y= 1/5x + 20 or 5y=x+100

Answer 3

I'm not sure if you've written the problem correctly. Anyway, here's the method you would use to solve:

First we can find the equation of line L by differentiating f(x) = x - (x^2/500), because the tangent line is the rate of change at point q. Thus...

df/dx = 1 - x/250, which can be writen as
L(x) = -x/250 +1 --> to put in y=mx+b form.

Next, we know one point that may be on the line L, namely (0,20). We will call this point P1. Now, we can see from L(x) that P1 is not on the tangent line, because when x = 0, L(x) = 1.

Answer 4

You sure you wrote the function correctly? If I'm reading it right, the graph starts at (0,0) and curves upward exponentially. So you start off with tangents with slope=0 and increase as the function moves to the right until they approach the vertical, and all those tangents cross the y-axis in the negative numbers; none of them pass through (0,20), so this question is not possible to solve.

Answer 5

i'm maths instructor sending you this education: a million) do no longer fix specified time, 2) interest is the motivating ingredient to earnings arithmetic. 3) very maximum mandatory factor to be suggested is "prepare Examples and workouts in basic terms. 4) do no longer omit any financial ruin. 5) in case you're a student of state board,do no longer learn something from out section the textual content textile e book. 6) Refer previous year questions. 6) if ISC, you may refer greater desirable books. 7) interest and problematical paintings with constructive might deliver you 100percentsuccesses