Finding Points of Intersection of a Parabolic Arch and a Hill

[physicsgal]

"a parabolic arch has an equation x^2 + 10y - 10 = 0. the arch is on a hill with equation y = 0.1x-1. (measurements are in metres).

"find the points of intersection"
for this i substituted y =0.1x - 1 into the equation x^2 + 10y - 10 = 0:
x^2 + 10(0.1x - 1) - 10 = 0
x^2 + x - 10 - 10 = 0
x^2 + x - 20 = 0
should this -20 be 0 instead? (i dunno)


and then used the quadratic equation so x = 4 or x = -5.
(putting those amounts into y =0.1x - 1), i have intersections of (4, -0.6) and (-5, -1.5). are these correct?

also, i don't understand how to draw this.

any help is appreciated!

~Amy

 

[physicsgal]

k, nevermind (figured it out) :)

~Amy

 

[Architeuthis Dux]

I would say that the points of intersection calculated using the quadratic equation are correct. However, it is important to note that these points represent the x-coordinate of the intersections. To find the y-coordinate, you will need to plug in the x-values into the equation y = 0.1x - 1.

As for drawing the parabolic arch and hill, you can plot the points of intersection and then sketch the curves of the equations. Alternatively, you can use a graphing calculator or software to plot the equations and see the intersection points visually. It may also help to label the axes and include a scale to accurately represent the measurements in meters.

I hope this helps! Let me know if you have any further questions.