Question

x² at the point P and intersects the x-axis at (-15,0) as shown in the Line 1 is tangent to the graph of y = x -- 120 figure above. a. Find the x-coordinates of point P. b. Write an equation for the line l. x2 c. If the line of symmetry for the curve y = x- 120 intersects line l at point R, what is the length of QR ?​

Answer 1

Explanation:

a) We know that y' = 1 - x/60. This will be the slope of our tangent. However, we also know this tangent has an x-intercept of (-15, 0).

This means that y - 0 = (1 - x/60)(x + 15).

From here you can find the intersection of this line with the parabola.

b) Find the equation of the line passing through (-15, 0) and pt P, nothing too hard

c) The axis of symmetry for the given curve is x=-b/2a = -(1)/(-1/60) = 60. So you need to find the point on line l where x=60, and find the distance from that pt to the vertex