Determining Quadratic Function for a Graph

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SUMMARY

The discussion focuses on determining the quadratic function for a graph with specific points: X-intercepts at (-3,0) and (5,0), a Y-intercept at (0,-30), and a local minimum/vertex at (1,-32). The quadratic function can be expressed in vertex form as y(x) = a(x-1)² - 32. To find the value of 'a', users can substitute the known points, such as y(5) = 0 or y(0) = -30, into the equation.

PREREQUISITES
  • Understanding of quadratic functions and their properties
  • Knowledge of vertex form of a parabola
  • Ability to solve equations for unknown variables
  • Familiarity with intercepts and their significance in graphing
NEXT STEPS
  • Practice deriving quadratic functions from given points
  • Learn about the significance of vertex form in graphing parabolas
  • Explore methods for finding the vertex and intercepts of quadratic functions
  • Study the impact of the coefficient 'a' on the shape of a parabola
USEFUL FOR

Students preparing for algebra quizzes, educators teaching quadratic functions, and anyone looking to enhance their understanding of graphing parabolas.

Sharpy1
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Hmm I'm having issues with an optional problem for some review for a quiz later tonight.

Determine the Quadratic Function for the Graph
the points labeled are X intercepts =(-3,0) (5,0); Y-Intercept = (0,-30), Local Min/Vertex = (1,-32)
View attachment 2420

It's the parabola looking problem at the bottom right, sorry I wasn't able to snap a good picture of it.

This is part of some optional review and I have a few hours before the quiz so any help would be appreciated. I'm honestly not even sure where to being other than putting "-32" at the end of the equation as the interpretation of the minimum point.
 

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Mathematics news on Phys.org
A parabola with vertex at $(x_0,y_0)$ has equation $y(x)=a(x-x_0)^2+y_0$ for some $a$. So, here the formula is $y(x)=a(x-1)^2-32$. Find $a$ from the fact that $y(5)=0$ or from $y(0)=-30$.
 

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