# A parabola has a y-intercept of 2 and passes through points (–2, –4) and (8, –14). Determine the vertex of the parabola. We are given that the points (-2,-4) and (8,-14) lie on a parabola with y-intercept of 2. We are asked to find the vertex.

We know that through three noncollinear points there lies exactly one parabolic function (there may be other parabolas through the same points, but they will not...

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We are given that the points (-2,-4) and (8,-14) lie on a parabola with y-intercept of 2. We are asked to find the vertex.

We know that through three noncollinear points there lies exactly one parabolic function (there may be other parabolas through the same points, but they will not represent functions.) One way to find the vertex is to find the equation of the function and then find the vertex:

In standard form the equation for a parabola is y = ax^2+bx+c

Using the given points we get the following:

For (0,2) we get a(0)^2+b(0)+c=2 so c=2.

Using this we get for (-2,-4) a(-2)^2+b(-2)+2=-4 or 4a-2b=-6 so 2a-b=-3

Then for (8,-14) we get a(8)^2+b(8)+2=-14 or 64a+8b=-16 so 8a+b=-2

We can now solve the system of equations:

2a-b=-3
8a+b=-2
---------
10a=-2 ==> a=-1/2 so b=2 and c=2

The equation, in standard form, for the parabola is y=-1/2x^2+2x+2

The x-coordinate for the vertex in standard form is given by x=-b/(2a) so we get x=(-2)/(-1)=2

Then the y-coordinate is y=-1/2(2)^2+2(2)+2=4

The vertex of the parabola is (2,4).

(1) From the standard form we can use completing the square to rewrite in vertex form:

y=-1/2x^2+2x+2=-1/2(x^2-4x+4)+2+2=-1/2(x-2)^2+4 so the vertex is (2,4)

(2) We can use matrices to solve the system of equations, or use substitution instead of linear combinations.

The graph: