Finding a parabola given two x intercepts

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SUMMARY

Given the x-intercepts (5,0) and (-1,0) of a parabola, it is impossible to determine a specific equation for the parabola using only these two points. A general parabola can be established, but to define a unique parabola, three points are necessary. The discussion confirms that there are infinitely many parabolas that can pass through these two intercepts, each varying in steepness and concavity. The vertex's y-coordinate cannot be determined solely from the line of symmetry without additional information.

PREREQUISITES
  • Understanding of quadratic equations in the form y = ax^2 + bx + c
  • Knowledge of x-intercepts and their significance in graphing parabolas
  • Familiarity with the concept of the vertex and line of symmetry in parabolas
  • Basic algebra skills for manipulating equations and solving for variables
NEXT STEPS
  • Research how to derive the vertex of a parabola from its x-intercepts
  • Learn about the general form of a quadratic equation and its parameters
  • Explore the concept of parabolas with varying coefficients and their graphical representations
  • Study the implications of having two roots on the characteristics of a parabola
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Mathematics students, educators, and anyone interested in understanding the properties and equations of parabolas, particularly in relation to their x-intercepts.

revoz
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If you are given (5,0) and (-1,0) as the two x intercepts of a parabola is it possible to find the equation of the parabola? I have tried using the vertex formula for the x co-ordinate which is x = 2 the line of symmetry and plugging in either of these co-ordinates into y = ax^2 + bx + c but have too many unknowns to solve for. Is this unsolvable with only this information?
 
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This is possible only if you are looking for a general parabola; NOT for a specific one. In fact, you need THREE points to establish a specific parabola.
 
Only being given the roots of the parabola isn't sufficient enough to compile a specific parabola from it. There are an infinite number of parabolas having those 2 roots, all having different "steepness" and concavity.

e.g.

y=(x-5)(x+1)
y=2(5-x)(3x+3)

etc.
 
Thanks. I suppose if I tried different values for a and b I would come up with different parabolas. Wasn't sure if there was some way with the line of symmetry to determine the y value of the vertex but I realize that there are different options with only two points. Thanks again.
 

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