Finding Points of Intersection of a Parabolic Arch and a Hill

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SUMMARY

The discussion focuses on finding the points of intersection between a parabolic arch defined by the equation x² + 10y - 10 = 0 and a linear hill represented by y = 0.1x - 1. The correct substitution leads to the quadratic equation x² + x - 20 = 0, yielding solutions x = 4 and x = -5. Substituting these x-values back into the hill's equation results in intersection points (4, -0.6) and (-5, -1.5), which are confirmed as accurate. The user also sought guidance on how to graph these equations.

PREREQUISITES
  • Understanding of quadratic equations and their solutions
  • Familiarity with graphing linear and quadratic functions
  • Knowledge of substitution methods in algebra
  • Basic skills in coordinate geometry
NEXT STEPS
  • Study the quadratic formula and its applications in solving equations
  • Learn how to graph parabolas and linear equations using graphing software
  • Explore the concept of intersections in algebraic functions
  • Investigate the use of graphing calculators or tools like Desmos for visual representation
USEFUL FOR

Students in algebra, mathematics educators, and anyone interested in understanding the intersection of quadratic and linear functions.

physicsgal
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"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
 
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k, nevermind (figured it out) :)

~Amy
 

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