Parabola passes through the points (0,4), (0,-1) and (6,1)

[HawKMX2004]

I have a few problems for a review section in my book for a test comming up soon. i am confused on a few problems and was wondering if some1 could help please?

1.) The Axis is horizontal; the parabola passes through the points (0,4), (0,-1) and (6,1)

I am very confused on this problem and 2 others exactly like it, but with different points. Could someone please help me get these started?

Thanks for all your help its appreciated

 

[Andrew Mason]

I have a few problems for a review section in my book for a test comming up soon. i am confused on a few problems and was wondering if some1 could help please?

1.) The Axis is horizontal; the parabola passes through the points (0,4), (0,-1) and (6,1)

What is the general equation for a parabola oriented like this? It should be x in terms of y^2 and three constants. So you need three equations to find those three constants. The points give you those three equations.

Use the general equation and plug in x = 0 and y=4, x=0, y =-1, and x=6, y=1 and you can figure out what the constants are.

AM

 

[wais]

A parabola is a U-shaped curve that can be described by the equation y = ax^2 + bx + c, where a, b, and c are constants. In order to find the equation of a parabola that passes through three given points, we can use a system of equations.

First, let's plug in the coordinates of the first point (0,4) into the equation y = ax^2 + bx + c:

4 = a(0)^2 + b(0) + c
4 = c

Next, we can plug in the coordinates of the second point (0,-1) into the equation:

-1 = a(0)^2 + b(0) + c
-1 = c

Now, we have two equations with two unknowns (a and b). We can solve this system of equations by substitution. Since we know that c = 4 and c = -1, we can substitute 4 for c in the second equation:

-1 = a(0)^2 + b(0) + 4
-1 = 4
-5 = 0

This is a contradiction, which means that there is no solution for this system of equations. This also means that the given points do not lie on a parabola with a horizontal axis.

To find the equation of a parabola with a horizontal axis that passes through the given points, we can use the general equation y = a(x-h)^2 + k, where (h,k) is the vertex of the parabola.

To find the vertex, we can use the formula h = -b/2a and k = c - b^2/4a. Since we know that c = 4 and b = 0, we can substitute these values into the formulas:

h = -0/2a = 0
k = 4 - 0^2/4a = 4

So, the vertex of the parabola is (0,4). We can now substitute these values into the general equation to find the specific equation of the parabola:

y = a(x-0)^2 + 4
y = ax^2 + 4

To find the value of a, we can use one of the given points. Let's use (6,1):

1 = a(6)^2 + 4
1 = 36a +