SOLUTION: please help.I posted these about 3 days ago,but haven't got a response.Can someone please help.thanks a parabola that opens to the right is a ? a. radicand b. function c.linear

Question 208475: please help.I posted these about 3 days ago,but haven't got a response.Can someone please help.thanks
a parabola that opens to the right is a ? a. radicand b. function c.linear equation d. relation
sketch the graphs using the roots to find the function and the turning point
(1).x= -8, x=4
(2)x= -3,x=3
use algebra to show that there are no common solutions.
(1)y = f(x)= x^2 + 4
y = x - 5
(2) y = f(x) = -2x^2
y = 4
find the common solution for each pair of linear equations
(1) 4y + 3x = 12
2x + 3y = 1
thank you for your time and help.I really appreciate it.

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
a parabola that opens to the right is a ?
a. radicand
b. function
c.linear equation
d. relation
---
Ans: It is a relation because there are two y values for each x-value
in the domain.
-----------------------------------------------
sketch the graphs using the roots to find the function and the turning point
(1).x= -8, x=4
Ans: f(x) = (x+8)(x-4) = x^2 +4x -32

Roots are x = -8 and 4
Turning point occurs at x = -b/2a = -4/2 = -2
Turning point y-value is f(-2) = (6)*(-6) = -36
Turning point is (-2,-36)
------------------------------------------------------------
---------------------------------------------------------
(2)x= -3,x=3
Ans: y = (x+3)(x-3) = x^2 -9

Roots are x = -3 and 3
----
Turning point occurs at x = -b/2a = -0/2 = 0
f(0) = -9
Turning point is (0,-9)
--------------------------------------------------------------
use algebra to show that there are no common solutions.
(1)y = x^2 + 4
y = x - 5
---
Substitute for "y" to get:
x^2+4 = x-5
x^2 - x + 9 = 0
---
The determinant is b^2-4ac = 1-4*9 = -35
Since the determinant is negative there are no Real Number solutions.
Therefore there are no points of intersection.
-------------------------------------------------------------
(2) y = -2x^2
y = 4
Substitute to get -2x2 = 4
2x^2+4 = 0
x^2 + 2 = 0
----
b^2 - 4ac = 0-4*2*1 is negative
No solutions and no points of intersection.
-------------------------------------------------
find the common solution for each pair of linear equations
(1) 4y + 3x = 12
....2x + 3y = 1
Ans:
Rearrange:
3x + 4y = 12
2x + 3y = 1
--------------------
Multiply thru the 1st Eq. by 2 and the 3nd equation by 3 to get:
6x + 8t = 24
6x + 9y = 3
-------------------
Subtract 1st from 2nd to get:
y = -21
----
Substitute into 2x + 3y = 1 to solve for "x":
2x + 3*-21 = 1
2x -63 = 1
2x = 64
x = 32
---
Solution: (32,-21)
===========================
Cheers,
Stan H.

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