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Proposition
If #u# is an odd integer, then the equation #x^2 + x - u = 0# has no solution that is an integer.

Proof
Suppose that there exists a integer solution #m# of the equation:

#x^2 + x - u = 0#

where #u# is an odd integer. We must examine the two possible cases:

#m# is odd; or
#m# is even.

First, let us consider the case where #m# is odd, then there exists an integer #k# such that:

# m = 2k + 1 #

Now, since #m# is a root of our equation, it must be that:

# m^2 + m - u = 0 #
# :. (2k + 1)^2 + (2k + 1) − u = 0 #
# :. (4k^2 + 4k + 1) + (2k + 1) − u = 0 #
# :. 4k^2 + 6k + 2 − u = 0 #
# :. u = 4k^2 + 6k + 2 #
# :. u = 2(2k^2 + 3k + 1) #

And we have a contradiction, as #2(2k^2 + 3k + 1)# is even, but #u# is odd.

Next, let us consider the case where #m# is even, then there exists an integer #k# such that:

# m = 2k #

Similarly, since #m# is a root of our equation, it must be that:

# m^2 + m - u = 0 #
# :. (2k)^2 + (2k) − u = 0 #
# :. 4k^2 + 2k − u = 0 #
# :. u = 4k^2 + 2k #
# :. u = 2(2k^2 + k) #

And, again, we have a contradiction, as #2(2k^2 + k)# is even, but #u# is odd.

So we have proved that there is no integer solution of the equation #x^2 + x - u = 0# where #u# is an odd integer.

Hence the proposition is proved. QED

Answer:

Vertex: #(4,18)#
Y-intercept: #(0, 2)#
X-interecpt: #(4+3sqrt2, 0)# and #(4-3sqrt2, 0)#

Explanation:

The vertex is pretty easy. We just need to complete the square .

First, the leading coefficient of the polynomial must be #1#, so we need to factor the #-1#. That leaves us with #y=-1(x^2-8x-2)#. Now, the purpose of completing the square is to find a constant that makes #x^2-8x# a perfect square. To that, we use this formula: #c=(1/2*b)^2# or #(1/2*8)^2#, which is #16#.

Now we know that we have to add #16# to make it a pefect square, but because we cannot just add something on one side of the equation, we need to "get rid of it" too. We could add #16# on both sides, or we can just add #16# and then subtract it immediately, which is the same thing. Either way works :)
#y=-1(x^2-8xcolor(green)(+16-16)-2)#
#y=-((x^2-8x+16)-16--2)#

#x^2-8x+16# is a perfect square, so let's symplify it

#y=-((x-4)^2-16-2)#
#y=-((x-4)^2-18)#

Now we just distribute the negative:
#y=-(x-4)^2+18#

The equation is now in vertex form.

It's easy to find the vertex from this point:
#y=-(x-color(red)(4))^2+color(purple)(18)#
#(color(red)(4), color(purple)(18))#.

Finding the #x#-interecpt means setting #y=0# and solving for #x#:

#0=-(x-4)^2+18#
#-18=-(x-4)^2#
#18=(x-4)^2#
#+-sqrt(18)=x-4#
#4+-sqrt(18)=x#
or #x=4+-3sqrt2#

Those are the exact values. If you want the estimated values, they're #x~~8.243# and #x~~-.0.243#

To find the #y#-intercept we just set #x=0# and solve for #y#:
#y=-(0-4)^2+18#
#y=-(-4)^2+18#
#y=-16+18#
#y=2#

Answer:

The inverse function is #f^-1(x)=lnx/ln3#

Explanation:

Let #y=3^x#

To calculate the inverse, interchange #x# and #y#,

#x=3^y#

Express #y# in terms of #x#

#lnx=ln(3^y)#

#lnx=yln3#

#y=lnx/ln3#

Therefore,

#f^-1(x)=lnx/ln3#

Verification :

#f(f^-1(x))=f(lnx/ln3)=3^(lnx/ln3)#

Let #y=3^(lnx/ln3)#

#=>#

#lny=lnx/ln3*ln3=lnx#

#y=x#

#f(f^-1(x))=x#

So, the functions #3^x# and #lnx/ln3# are inverses

Answer:

Use #y"-intercept"# and #x"-intercept"# or any two points on the line.

Explanation:

Looking at the equation, it is apparent that it would follow a linear model as there are no exponents or powers (eg #x^2#)

Then what we can do if find to points on the graph and draw a line through these points. The easiest points to find are usually the #y"-intercept"# and #x"-intercept"#

The #y"-intercept"# is when #x=0#, so to find that point, make #x=0# in the equation, and solve for #y#

#x=0#

#=>-2(0)+y=2#

#=>y=2#

Hence, when #x=0#, #y=2#. This represents the point #(0,2)#. So we should plot this point:

Geogebra.com

Then we calculate the other point, the #x"-intercept"#

The #x"-intercept"# is when #y=0#, so to find that point, make #y=0# in the equation, and solve for #x#

#y=0#

#=>-2x+(0)=2#

#=>-2x=2#

#=>x=2/-2#

#=>x=-1#

Hence, when #y=0#, #x=-1#. This represents the point #(-1,0)#. So we should plot this point with the other point:

Geogebra.com

Now join the two points with a straight line:

Geogebra.com

And you have the graph of your equation

You can use any two poinst on the line though. The only difference is that instead of finding when #x=0# or #y=0#, you would find when #x="another number like "3# or when #y="another number like "4#

It is easier to find the #y"-intercept"# and #x"-intercept"# but if you want to challenge yourself, find two points other than the #y"-intercept"# and #x"-intercept"#

Answer:

Range of a #y = f(x) # is a set of all the outputs corresponding to their input in the domain

Explanation:

there are many ways to find range
1. by constructing graph :
example
#y = sinx#
Domain :: #(-oo, oo)#
Range :: #(-1,1)#
because we get values between -1 and 1 (both included)
graph{sinx [-10, 10, -2, 2]}

  1. solving #x = f(y) #
    example :

#f(x) = (2x + 1) /(x -1)#
Domain : #R -{1}#
enter image source here

Now , forming equation such as # x = f(y)#

#x = (y + 3)/(y -2)#
graph{(x +3)/(x -2) [-41.1, 41.08, -8.22, 8.22]}
Range : #R -{2}#

  1. if you know that the function is continuous in domain then
    then you can find max and min values of y

Ex . Range of #f(x) = x^3 - 6x^2+11x-6# where x lies b/w #(0 , 6)#
Ans. here, function is continuous in domain x : #(0,6)#
and # y_max = f(6) = 60 and y_min = f(0) = -6#
So, Range : #(-6,60)#

Another Example which is important is when
#f(x) = x^2 - 3x +2# where x lies b/w #(-oo , oo)#
Here #f(x)# is continuous in the domain and
#y_max = f(-oo) = oo # but,
#y_min = f(3/2) = -1/4#
So,
Range is #[-1/4,oo)#

Answer:

Condition :: #m_1m_2 = -1#
Slope :: #1/3#

Explanation:

Suppose that the slope of the 2 lines be #m_1# and #m_2# respectively
and angle between line and #x# -axis be #theta_1# and #theta_2# respectively.

So, #m_1 = tan(theta_1) and m_2 = tan(theta_2)#
we also know that #theta_1 - theta_2 = pi/2#

then ,
#tan(theta_1 - theta_2) = oo#

but,

#tan(theta_1 - theta_2) = (tantheta_1 - tantheta_2)/(1 + tantheta_1tantheta_2)#

#oo = (m_1 - m_2)/(1 + m_1m_2)#

this can happen when

#m_1m_2 = -1#
(this condition is very important )

substituting #m_1 = -9/3 = -3 # (from the equation)
we get #m_2 =1/3#

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