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the vertex of the parabola below is at the point (4, -1). Which could be this parabola's equation?

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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Kenneth G. | Experienced Tutor of Mathematics and StatisticsExperienced Tutor of Mathematics and Sta...
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) = ax2 + 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) = ax2 -8ax + 16a - 1 = a(x2-8x+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.  
Steve S. | Tutoring in Precalculus, Trig, and Differential CalculusTutoring in Precalculus, Trig, and Diffe...
5.0 5.0 (3 lesson ratings) (3)
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.
Deanna L. | Electrical engineering major and music lover with MIT degreeElectrical engineering major and music l...
4.9 4.9 (120 lesson ratings) (120)
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.


That is how the quadratic formula derives. Factoring with completing the square of aX2 + bx +c.
Parviz F. | Mathematics professor at Community CollegesMathematics professor at Community Colle...
4.8 4.8 (4 lesson ratings) (4)
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
   For this case:
    f( x) = K( X - 4 ) ^2 - 1
            = k( X^2 -8X + 16 -1=
             = K ( X2- 8X +15) , have to have coordinate of another point of the curve to determine value of K,