Line l is tangent to the graph of

y= x - x²/500 at point Q & it crosses the y-axis at (0, 20)

a). Find the x-coordinate of point Q
b. Write an equation for line l
c) Suppose the graph of y=x-x^2/500, where x and y are measured in feet, represents a hill. There is a 50-foot tree growing vertically at the top of the hill. Does a spotlight at point P (point P is where line l crosses the x-axis) directed along the line l shine on any part of the tree? Show the work that leads to your conclusion.

a) To find the x-coordinate of point Q, we need to find the point where line l is tangent to the graph. To do this, we can start by finding the derivative of the function y = x - x²/500.

The derivative of y = x - x²/500 is dy/dx = 1 - (2x)/500.

To find the x-coordinate of the tangent point Q, we can set dy/dx equal to 0 and solve for x:

0 = 1 - (2x)/500
(2x)/500 = 1
2x = 500
x = 250

So the x-coordinate of point Q is 250.

b) Now that we have the x-coordinate of point Q, we can plug it back into the equation y = x - x²/500 to find the corresponding y-coordinate:

y = 250 - (250)²/500
y = 250 - 62.5
y = 187.5

So the coordinates of point Q are (250, 187.5).

To write an equation for line l, we can use the point-slope form of a line, which is y - y₁ = m(x - x₁), where (x₁, y₁) is a point on the line and m is the slope.

Using point Q (250, 187.5), we already know the x and y coordinates, so we just need to find the slope. The slope can be found using the derivative dy/dx:

m = dy/dx at Q = 1 - (2(250))/500
m = 1 - 1
m = 0

So the equation for line l is y - 187.5 = 0(x - 250), which simplifies to y = 187.5.

c) Since the equation of the line l is y = 187.5, it means that the line is parallel to the x-axis and does not intersect the tree. Therefore, the spotlight at point P, where line l crosses the x-axis, does not shine on any part of the tree.

a) To find the x-coordinate of point Q, we need to find the point where line l is tangent to the graph of y = x - x^2/500.

To do this, we need to find the derivative of the function y = x - x^2/500 and set it equal to the slope of line l. The derivative of y = x - x^2/500 is:

dy/dx = 1 - (2x/500) = 1 - (x/250)

Since line l is tangent to the graph at point Q, its slope should be equal to the derivative of the function at that point. Let's assume the x-coordinate of point Q is q.

Therefore, the slope of line l is:

1 - (q/250)

Now, we know that line l crosses the y-axis at the point (0, 20). The equation of a line in slope-intercept form is y = mx + c, where m is the slope and c is the y-intercept.

Plugging in the slope and the point (0, 20), we have:

20 = (1 - (q/250)) * 0 + c
20 = c

So, the y-intercept of line l is 20.

b) Now that we have the slope and the y-intercept, we can write the equation of line l as:

y = (1 - (q/250)) * x + 20

b) The equation for line l is y = (1 - (q/250)) * x + 20.

c) To determine if the spotlight at point P shines on any part of the tree, we need to find the y-coordinate of the point where line l crosses the x-axis.

When a point lies on the x-axis, the y-coordinate is 0. So, we set y = 0 in the equation of line l and solve for x:

0 = (1 - (q/250)) * x + 20
(q/250) * x = -20
x = -20 * (250/q)

We know that the tree is at a height of 50 feet. If the spotlight shines on any part of the tree, the y-coordinate of point P should be greater than or equal to 50.

To determine if this is the case, we substitute x = -20 * (250/q) into the equation of the graph y = x - x^2/500:

y = (-20 * (250/q)) - ((-20 * (250/q))^2 / 500)

Simplifying this equation will give us the height of the point P. If the height is greater than or equal to 50, the spotlight shines on the tree.

To find the answers to the questions, let's break it down step by step.

a) Find the x-coordinate of point Q:
To find the x-coordinate of point Q, we need to find the point where the line is tangent to the graph. Since line l is tangent to the graph, it means that at point Q, the slope of the line is equal to the slope of the graph. The slope of the graph can be found by taking the derivative of the given equation.

y = x - x^2/500

To find the derivative, we differentiate each term:
dy/dx = 1 - (2x)/500

Since line l is tangent to the graph at point Q, its slope is equal to the derivative at that point. Let's assume that the x-coordinate of point Q is q.

Slope of line l = derivative at point Q
=> 1 = 1 - (2q)/500

Simplifying the equation:
(2q)/500 = 0
2q = 0
q = 0

Therefore, the x-coordinate of point Q is 0.

b) Write an equation for line l:
We know that line l crosses the y-axis at (0, 20). So, we have a point (0, 20) on the line. Additionally, we know that the line is tangent to the graph at point Q, which we found in part a to be (0, 0). Using these two points, we can find the slope of the line and then write its equation.

Slope = (Change in y) / (Change in x)
= (20 - 0) / (0 - 0)
= 20 / 0 (which is undefined)

Since the slope is undefined, we can conclude that line l is a vertical line passing through the point (0, 20). Therefore, the equation for line l is x = 0.

c) To determine if the spotlight at point P shines on any part of the tree, we need to check if the y-coordinate of the tree is less than or equal to zero, as the tree lies above the x-axis and the line l crosses the x-axis at point P.

The y-coordinate of the tree is 50 feet, and we need to determine if it is less than or equal to zero.

y-coordinate of the tree < 0?
50 < 0

Since the inequality is not true (as 50 is greater than zero), we can conclude that the spotlight at point P does not shine on any part of the tree.

dy/dx = slope = 1 - x/250

y = m x + 20

at tangent point
y = (1-x/250) x + 20 = x - x^2/500

solve that for x