Question 321729: Given a parabola with intersection points (-1,0), (2,0) and (0,-16) and a line which passes through (-5,-10) and (4,30), find an equation for the line and the parabola and the points of intersection between the two graphs if any.
Found 2 solutions by Fombitz, solver91311:
Answer by Fombitz(32388)   (Show Source):
You can put this solution on YOUR website! The general equation for a parabola is, 
Plugging in the points you get a system of equations,
1.
2.
3.
Substitute eq. 3 into eq. 1 and eq. 2,
4.
5.
Multiply eq. 4 by 2 and add to eq. 5 to solve for a.
Then from eq. 4,
The parabola is then,
.
.
.
For the line, calculate the slope,
Using the point slope form of a line,  , use either point to solve for b.
The line is then,
To find the intersection points set the two function equal to each other and solve for x.
.
.
.
.
.
.
Use the quadratic formula,
Now that you have the x values, use either equation to solve for y.
Answer by solver91311(24713)  (Show Source):
You can put this solution on YOUR website!
The  coordinates of the -intercepts of a polynomial function are the zeros of the function. The Fundamental Theorem of Algebra says that a degree  polynomial equation has roots, hence the described parabola, being a 2nd degree polynomial, has exactly 2 roots. If  is a zero of a polynomial, then  is a factor of the polynomial.
The zeros are -1 and 2, hence the factors are  and  . Therefore, the general form of the polynomial is:
\ =\ k(x\ +\ 1)(x\ -\ 2)\ =\ k\left(x^2\ -\ x\ -\ 2\right))
Given the initial condition (based on the coordinates of the  -intercept, ) ), allows us to calculate 
\ =\ kx^2\ -\ kx\ -\ k2)
Since
\ =\ -16)
^2\ -\ k(0)\ -\ k2\ =\ -16)
And the polynomial is
\ =\ 8x^2\ -\ 8x\ -\ 16)
An equation of the line that passes through ) and ) is given by the two-point form:
(x\ -\ x_1) )
(x\ +\ 5) )
Set the two RHSs equal:
Real roots of the above equation, if any exist, will be the coordinates of the ordered pair(s) that represent the solution set of the system of equations.
But (146)\ =\ -41024\ <\ 0)
So the roots are a conjugate pair of complex numbers with a non-zero imaginary part. Therefore there are no real roots, and therefore no points of intersection of the two graphs.
John
|
|
|
