Suppose the number is x, its reciprocal is 1/x
x+1/x=13/6
(x^2+1)/x=13/6
6(x^2+1)=13x
6x^2-13x+6=0
(3x-2)(2x-3)=0
x=2/3 or x=3/2
If the coefficient of x^2 is negative, the graph will be n shaped and curve down instead of like a u shape if it was positive. If the vertex is below the x axis and curves down, it won’t pass the x axis, Tia is right.
It would be A or B
but not D,C
Answer: 0
35X +14=3X +14
35X = 3X
35X-3X=0
32X = 0 divide both sides by 32
x=0
So, pretend this is your x-axis and y-axis:
I
I
(-2,7) • I
I
I • (2, 5)
I
I
I
I
_________________I____________________
I
I
I
TO GET FROM POINT (-2, 7) TO POINT (2, 5), WE MOVE DOWN 2 AND OVER 4, SO THE SLOPE IS -1/2. IF WE FOLLOW THAT SLOPE AND MOVE DOWN 1 AND OVER 2 FROM THE FIRST POINT OF (-2, 7), WE WILL LAND ON A POINT LOCATED AT (0, 6), WHICH WOULD BE THE "Y-INTERCEPT". WE WERE JUST ABLE TO CALCULATE THE SLOPE OF THE LINE AND THEN USE THE SLOPE TO FIND THE INTERCEPT. SO, THE "SLOPE-INTERCEPT" FORM OF THE EQUATION FOR THIS LINE IS:
y = -1/2x + 6
TO RE-WRITE THIS IN STANDARD FORM, WE JUST WANT TO MOVE THE X VARIABLE OVER TO THE LEFT WITH THE Y VARIABLE, SO:
y = -1/2x + 6
+1/2x + 1/2x
1/2x + y = 6 .... and that is your answer!
