Answer:
Shift the graph up 3 units.
Let's to the first example:
f(x) = x^2 + 9x + 20
Ussing the formula of basckara
a = 1
b = 9
c = 20
Delta = b^2 - 4ac
Delta = 9^2 - 4.(1).(20)
Delta = 81 - 80
Delta = 1
x = [ -b +/- √(Delta) ]/2a
Replacing the data:
x = [ -9 +/- √1 ]/2
x' = (-9 -1)/2 <=> - 5
Or
x" = (-9+1)/2 <=> - 4
_______________
Already the second example:
f(x) = x^2 -4x -60
Ussing the formula of basckara again
a = 1
b = -4
c = -60
Delta = b^2 -4ac
Delta = (-4)^2 -4.(1).(-60)
Delta = 16 + 240
Delta = 256
Then, following:
x = [ -b +/- √(Delta)]/2a
Replacing the information
x = [ -(-4) +/- √256 ]/2
x = [ 4 +/- 16]/2
x' = (4-16)/2 <=> -6
Or
x" = (4+16)/2 <=> 10
______________
Now we are going to the 3 example
x^2 + 24 = 14x
Isolating 14x , but changing the sinal positive to negative
x^2 - 14x + 24 = 0
Now we can to apply the formula of basckara
a = 1
b = -14
c = 24
Delta = b^2 -4ac
Delta = (-14)^2 -4.(1).(24)
Delta = 196 - 96
Delta = 100
Then we stayed with:
x = [ -b +/- √Delta ]/2a
x = [ -(-14) +/- √100 ]/2
We wiil have two possibilities
x' = ( 14 -10)/2 <=> 2
Or
x" = (14 +10)/2 <=> 12
________________
To the last example will be the same thing.
f(x) = x^2 - x -72
a = 1
b = -1
c = -72
Delta = b^2 -4ac
Delta = (-1)^2 -4(1).(-72)
Delta = 1 + 288
Delta = 289
Then we are going to stay:
x = [ -b +/- √Delta]/2a
x = [ -(-1) +/- √289]/2
x = ( 1 +/- 17)/2
We will have two roots
That's :
x = (1 - 17)/2 <=> -8
Or
x = (1+17)/2 <=> 9
Well, this would be your answers.
Answers:
1) Given
2) angle 2 ** see note below
3) angle 3 ** see note below
4) converse of alternate exterior angle theorem
note: you can swap the answers for 2 and 3 and it doesn't matter
========================================================
Explanations:
1) This is given so we just simply state "given". It seems silly to repeat what is given, but this is how you start any geometry proof.
2 & 3) The answers here are angle 2 and angle 3 because they are both interior angles (on the inside of the parallel lines m and L) and they are on alternate sides of the transversal line q. So they are both alternate interior angles and are congruent due to line L parallel to line m (alternate interior angle theorem)
4) If you have a pair of parallel lines, then the alternate exterior angle theorem says that alternate exterior angles are congruent. Going in reverse, the converse of this theorem says that having a pair of congruent alternate exterior angles (angle 1, angle 2) leads to the lines being parallel (p and q).
Answer:
X^2-2X+1=0
Step-by-step explanation:
given,2X-X^2=1
or,-(X^2-2X)=1
or,X^2-2X=-1
or,X^2-2X+1=0 ,which is the req.equation in standard form.
