Find the equation for a parabola with given vertex and point.

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To find the equation of a parabola with vertex (-5, 5) that passes through the point (-3, 17), the quadratic form ax^2 + bx + c is used. By substituting the vertex and the given point into the equations, two equations are formed: 5 = 25a - 5b + c and 17 = 9a - 3b + c. The vertex condition indicates that -5 = -b/2a, allowing for the determination of coefficients. The resulting equation is y = 3(x + 5)^2 + 5, which correctly identifies the vertex and satisfies the point (-3, 17). This method effectively demonstrates how to derive the equation of a parabola from given parameters.
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Vertex (-5,5) point (-3,17)

Equation for the parabola with the given vertex that passes through the given point.

I am stumped! Please help :bugeye:
 
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well, look at this
you have a quadratic of the form ax^2+bx+c
so you have two points
so plug in y and x in for those two points:

5=25a-5b+c
and
17=9a-3b+c

but they also tell you that (-5, 5) is the vertex
which means that there, the solution only has one root; meaning that the x coordinate is equal to -b/2a
so then you have
-5=-b/2a

i think you should be able to solve for this now. perhapsably.
 
Using what is provided, I have come up with:

y=3(x+5)^2+5
 
TonyC said:
Using what is provided, I have come up with:

y=3(x+5)^2+5

Great. This is always the most insightful way to write the equation of a parabola.
As a check, you can easily see the vertex is at (-5,5). Putting x=-3 gives y=3(2^2)+5=17.
 
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