^{2}+ bx + c.

^{2}-8ax + 16a - 1 = a(x

^{2}-8x+16) -1. This is the FAMILY of parabolas with vertex (4,-1).

I don't understand the question. What would the equation be for this vertex of the parabola? I'm doing Conic sections, and I don't understand the question.

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The vertex of the parabola is at (4,-1). Knowing the vertex alone will not determine a parabola. Assume the parabola's equation is f(x) = ax^{2} + bx + c.

The x-coordinate of the vertex is x = -b/2a = 4, so that b = -8a.

(4,-1) is on the parabola, so f(4) = -1 = 16a + 4b + c;

substituting, b = -8a, -1 = 16a -32a + c, so that c = 16a -1.

So f(x) = ax^{2} -8ax + 16a - 1 = a(x^{2}-8x+16) -1. This is the FAMILY of parabolas with vertex (4,-1).

The limit of the family as a approaches 0 is the horizontal line y = -1. As a increases to infinity the parabola becomes steeper, until a approaches infinity, the parabola approaches the vertical line segment at x = 4 where y is in (-1,infinity). As a decreases to -infinity, the parabola approaches the vertical like at x =4 where y is in the interval (-infinity, -1).

If we specify any point in the x-y plane where x is not equal to 4, there is a unique parabola in this family that passes through that point.

It's easy to see that y = a x^2 is a parabola with vertex at (0,0).

If it's translated by the vector <h,k> the equation becomes

(y - k) = a (x - h)^2, or y = a (x - h)^2 + k, and the vertex becomes (h,k).

(See sks23cu dot net slash MT slash Files slash Transformations slash sks23cuTransformations dot pdf)

If (h,k) = (4,-1), then the equation, by substitution, is y = a (x - 4)^2 - 1.

To find the value of "a" another point on the parabola must be used. Let's assume we know (5, -3) is a point on the parabola. Then:

-3 = a (5 - 4)^2 - 1 => -2 = a (1) => y = -2 (x - 4)^2 - 1.

The other way to approach this is to follow the general form of a parabola. Start with something simple.

y=x^2 is a parabola with a vertex at (0,0) right?

Well y=x^2 -1 would be a vertex at (0,-1). Getting close.

Now we need the vertex at x=4. So there's a neat trick here. I can prove it in a separate comment but it's easier to look at it logically. If y=(x-a)^2-1 would produce a vertex at y=-1, then (x-a)^2 would be 0. That means to have a vertex at x=4 would mean a=4.

So the equation could be y=(x-4)^2-1.

If you multiply it out, you'll find that this is x^2-8x+16-1 which is just what Parviz derived.

Parviz F. | Mathematics professor at Community CollegesMathematics professor at Community Colle...

Vertex of a parabola with equation:

f(x) = aX^2 + bX +c

is at Coordinate( ( -b/2a,b^2 - 4ac )

4a^2 )

f( x ) = k ( X - b/2a) ^2 - b^2 - 4ac

4a^2

For this case:

f( x) = K( X - 4 ) ^2 - 1

= k( X^2 -8X + 16 -1=

= K ( X^{2}- 8X +15) , have to have coordinate of another point of the curve to determine value of K,

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^{2 }+ bx +c.