Why is 3 = a x (1)^2 x (-3)^2 in this step of the solution?

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Homework Help Overview

The discussion revolves around a mathematical expression involving a variable 'a' and the evaluation of a function at specific points derived from a graph. The subject area includes algebra and graph interpretation.

Discussion Character

  • Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • Participants explore the evaluation of the expression by substituting specific values for 'x' and 'y' based on graphical information. There is a focus on understanding the implications of these substitutions and the significance of roots of multiplicity in the context of the graph.

Discussion Status

The discussion is progressing with some participants expressing understanding after clarifications. There is an exploration of the relationship between the graph and the algebraic expressions, indicating a productive exchange of ideas.

Contextual Notes

Participants reference specific values derived from the graph, such as x=0 and y=3, and discuss the nature of roots in relation to the function's behavior. The mention of roots of multiplicity suggests a deeper exploration of the function's characteristics.

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Homework Statement
I understand everything before and after this line but not sure how they actually came to that conclusion, many thanks.
Relevant Equations
ax^4 + bx^3 + cx^2 + dx + 3
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1705136497876.png
 
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Because they stated that x=0, therefor a(x+1)^2 becomes a(0+1)^2 = a(1)^2 and (x-3)^2 becomes (0-3)^2 = (-3)^2.

Note that the values of x=0 and y=3 come directly from the graph. You can see that the function (the squiggly line on the graph) is at a y-value of 3 when it crosses the y-axis, which corresponds to an x-value of 0. Therefor x=0 and y=3, which you just plug into the equation.
 
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Ahhhh I see it now, many thanks!
 
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For those that are wondering why -1 and 3 are roots of multiplicity 2, it is because from the graph we can see that they are roots of y(x)=0 but also they are local minima hence also roots of y'(x)=0.
 

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