Hi there!
a.
We can find the x-coordinate by deriving an expression for the line and arc, then setting the two equal.
We are given that the graph has the equation:
[tex]y = x - \frac{x^2}{120}[/tex]
Differentiate to solve for the slope:
[tex]y' = 1 - \frac{x}{60}[/tex]
This is the slope of the line at 'p'. We are given that the line passes through (-15, 0), so we can use the point-slope formula:
[tex]y - y_1 = m(x - x_1)\\\\y - 0 = (1 - \frac{x}{60})(x - (-15))\\\\y = (1 - \frac{x}{60})(x + 15)[/tex]
Since this line intersects at 'p', we can set this equation along with the equation for the arc equal to each other to solve for 'x'.
[tex](1 - \frac{x}{60})(x + 15) = x - \frac{x^2}{120}\\[/tex]
Solve using a graphing utility.
[tex]x = 30[/tex]
b.
We can write an equation for line 'l' by solving for the slope at the 'p' value.
[tex]m = 1 - \frac{30}{60} = 1 - 0.5 = 0.5[/tex]
Now, use the point-slope formula with this y-value and the x-axis intersection coordinates.
[tex]y - 0 = 0.5(x + 15)\\\\y = 0.5x + 7.5[/tex]
c.
We can plug x = 60 into both function equations to find the distance.
For the arc:
[tex]f(60) = 60 - \frac{60^2}{120} = 30[/tex]
For the line:
[tex]f(60) = .5(60) + 7.5 = 37.5[/tex]
Subtract to find the distance. (QR)
[tex]37.5 - 30 = 7.5[/tex]
