Find Equation for the Parabola- Help, Please?

[dmbeluke]

Each pair of the following three lines cross at a point. Those points are the y-intercept, one of the x-intercepts, and the vertex of a parabola. Can you please explain to me how to find an equation for the parabola? And the other x- intercept?

y+8x=32
y+5x=32
y+3x=12

I have been trying everything I can think of. I think they cross at 0. And I know the equation has to be squared, but I'm so lost on this one, I don't know what else to do. Any advice would be appriciated. Thank you.

 

[arildno]

Step 1:
Determine the 3 pairs of (x,y)-values that solves the 3 pairwise systems of linear equations.
Let's call these:
[tex] {r}_{1}=(x_{1},y_{1}),
{r}_{2}=(x_{2},y_{2}),
{r}_{3}=(x_{3},y_{3})
[/tex]

Step 2.
The general equation for a parabola when the y-coordinate is a function of the x-coordinate is:
[tex]
y=ax^{2}+bx+c[/tex], where a, b,c are constants.

You do know something of the parabola you're after, namely that the 3 pairs [tex] r_{1},r_{2},r_{3}[/tex] are on it.
How can this information be used in order to determine a,b,c?

 

[M98Ranger]

To find the equation for a parabola, we first need to understand the general form of a parabola. A parabola can be written in the form y = ax^2 + bx + c, where a, b, and c are constants. In this case, we have three equations with three unknowns (a, b, and c), so we can solve for them using a system of equations.

First, let's rearrange the equations to put them in the form y = ax^2 + bx + c:

y = 32 - 8x
y = 32 - 5x
y = 12 - 3x

Now, we can set up a system of equations by equating the coefficients of x^2, x, and the constants on both sides:

32 - 8x = a + bx + c
32 - 5x = a + bx + c
12 - 3x = a + bx + c

We can solve this system using any method we are comfortable with, such as substitution or elimination. For simplicity, let's use substitution:

From the first equation, we get:

a = 32 - 8x - bx - c

Substituting this into the second equation, we get:

32 - 5x = (32 - 8x - bx - c) + bx + c

Simplifying, we get:

32 - 5x = 32 - 8x

Solving for x, we get:

x = 4

Now, we can substitute this value of x into any of the original equations to solve for a, b, and c. Let's use the first equation:

y = 32 - 8(4) = 0

So, the y-intercept of the parabola is (0,0). This confirms what you mentioned in your post.

To find the other x-intercept, we can use the fact that the x-intercepts are symmetric about the vertex. Therefore, the other x-intercept will be at x = -4. Substituting this into any of the equations, we get the y-coordinate of the vertex:

y = 32 - 8(-4) = 64

So, the vertex of the parabola is at (-4,64).

Now, we can substitute these values into the general form of a parabola to get the equation:

y