MHB Determining Quadratic Function for a Graph

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To determine the quadratic function for the given graph, start with the vertex form of a parabola: y(x) = a(x - 1)² - 32. The vertex is at (1, -32), which establishes the minimum point. To find the value of 'a', use the x-intercept at (5, 0) or the y-intercept at (0, -30) to create equations. Substituting these points into the vertex form will allow for the calculation of 'a', completing the quadratic function. This method effectively utilizes the vertex and intercepts to derive the equation of the parabola.
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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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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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