A parabola indicates that it is a quadratic relation.
1. We will get the equation first in standard form. - substitution and elimination will be used here.
2. We will convert the equation from standard form to a vertex form. - completing the square method will be used here.
In a standard form of any quadratic relation-- y=ax^2+bx+c, the c value is the value of the y-intercept. Or you can also think of y-intercept as a coordinate of (0,2), and sub in x=0 and y=2, to isolate for c=2.
Our equation can be narrowed down to: y=ax^2+bx+2
With the two points that are given, you can sub in the points as x and y values into our narrowed down equation.
first point (-2, -4) => x=-2 and y=-4
y=ax^2+bx+2
-4=a(-2)^2+b(-2)+2 <- from here, simplify as much as you can
-6=a(4)+b(-2)
-6=4a-2b
second point
(8, -14) => x=8 and y=-14
y=ax^2+bx+2
-14=a(8)^2+b(8)+2 <- from here, simplify as much as you can
-16=a(64)+b(8)
-16=64a+8b
Taking the two equations, you can use either substitution or elimination to solve for your values of a and b
I am going to show you using elimination
from the equation obtained using the first point,
-6=4a-2b -> multiply this by 4
-24=16a-8b
Take
-24=16a-8b -> from first point
-16=64a+8b -> from second point
add the two equations
-24=16a-8b
-16=64a+8b
---------------
-40=80a -> now isolate for a
a=-0.5
using a, take any equation from the two points and sub this value in to obtain the value of b
-24=16a-8b
-24=16(-0.5)-8b
8b=16
b=2
Your quadratic equation in standard form is:
y=-0.5x^2+2x+2
You can use completing the squares method to obtain your vertex.
y=-0.5(x^2-4x)+2
y=-0.5(x^2-4x+4-4)+2
y=-0.5((x-2)^2 - 4)+2
y=-0.5(x-2)^2 + 2 +2
y=-0.5(x-2)^2 +4
Your vertex values are (h,k) from the vertex form y=a(x-h)^2+k
Your vertex is (2, 4).
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=-38a+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:
http://mathworld.wolfram.com/Parabola.html
http://mathworld.wolfram.com/QuadraticEquation.html
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