HELP with FINDING EQUATIONS AND INTERSECTION PTS?

Question

ok.. i need to find 2 quadratic functions (one opening up and one opening down) with these x intercepts :

(-1,0) (3,0)

and

(-3,0) (-1/2,0)


Also, intersection points for

y=9- x^2
y=x+3

and

y=x^3+2x-1
y=-2x+15

Please explain how to do them!! I've tried so hard to figure these out. Thanks.

Answer 1

The quadratic question: The x-intercepts of a parabola (the graph for a quadratic equation) tell you what the zeros of the quadratic are, and thus what the _factor_ making up the quadratic are. If -1 is an x-intercept, then it is also a zero of the quadratic polynomial, and it comes from a factor of (x + 1). Likewise, (3,0) tells you that there's a factor of (x - 3). Then the quadratic you're looking for is (x + 1)(x - 3) = x^2 - 2x - 3.

[To see why this works, start with x^2 - 2x - 3. Suppose you are asked to find the x-intercepts. You do that by setting the function equal to zero:
x^2 - 2x - 3 = 0. To solve, factor:
(x - 3)(x + 1) = 0. Set each factor equal to 0:
x - 3 = 0, and x + 1 = 0.
Then x = 3 and x = -1.
The reverse of this process is how your question gets answered.]

You can multiply this expression by any constant, and it will have the same x-intercepts. For example, multiplying by -1 will give you the expression for a parabola that opens downward.

On finding the intersection points:

Fortunately you have a bunch of equations where y is an explicit function of x. For each question, set the two equations equal to each other: for example,
y = 9 - x^2
y = x + 3
becomes
9 - x^2 = x + 3.

Then solve. The idea is that the point of intersection is a point (an x-coord and y-coord) that solves both equations at the same time. If a certain y-coordinate has to solve both, then it's the same value in each equation. If it's the same value, then the y in the first equation must equal the y in the second equation. Then the things that y is equal to must also be equal.

Answer 2

(-1,0) and (3,0)
(x + 1)(x - 3)
x^2 - 3x + x - 3
x^2 - 2x - 3

ANS : x^2 - 2x - 3 and -x^2 + 2x + 3

--------------------------

(-3,0) and ((-1/2),0)
(x + 3)(2x + 1)
2x^2 + x + 6x + 3
2x^2 + 7x + 3

ANS : 2x^2 + 7x + 3 and -2x^2 - 7x - 3

-----------------------------------------------------------------------------

y = 9 - x^2
y = x + 3

x + 3 = 9 - x^2
x^2 + x - 6 = 0
(x + 3)(x - 2) = 0
x = -3 or 2

ANS : (-3,0) and (2,5)

-------------------------------------

y = x^3 + 2x - 1
y = -2x + 15

-2x + 15 = x^3 + 2x - 1
x^3 + 4x - 16 = 0
(x - 2)(x^2 + 2x + 8) = 0

x = 2, -1 + isqrt(7) or -1 - isqrt(7)