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Question 204503:
Answer by jim_thompson5910(35256) About Me  (Show Source):
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# 1
f%28x%29=x%5E2%2B3 Start with the given function


f%28x%2Bh%29=%28x%2Bh%29%5E2%2B3 Replace each "x" with "x+h"


f%28x%2Bh%29=x%5E2%2B2xh%2Bh%5E2%2B3 FOIL (ie expand)

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%28f%28x%2Bh%29-f%28x%29%29%2Fh Move onto the given difference quotient.


%28x%5E2%2B2xh%2Bh%5E2%2B3-%28x%5E2%2B3%29%29%2Fh Plug in f%28x%2Bh%29=x%5E2%2B2xh%2Bh%5E2%2B3 and f%28x%29=x%5E2%2B3


%28x%5E2%2B2xh%2Bh%5E2%2B3-x%5E2-3%29%2Fh Distribute


%282xh%2Bh%5E2%29%2Fh Combine like terms.


%28h%282x%2Bh%29%29%2Fh Factor out the GCF "h" from the numerator.


%28cross%28h%29%282x%2Bh%29%29%2Fcross%28h%29 Cancel out the common terms.


2x%2Bh Simplify


So %28f%28x%2Bh%29-f%28x%29%29%2Fh=2x%2Bh when f%28x%29=x%5E2%2B3


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# 2

Take note that when x=-3 it is less than zero. Because f%28x%29=-x when x%3C0, this means that we simply plug in x=-3 to get: f%28-3%29=-%28-3%29=3.

So f%28-3%29=3

Also, when x=2 it is greater than zero. Since f%28x%29=x%5E3 when x%3E=0, we just plug in x=2 to get: f%282%29=%282%29%5E3=8.


So f%282%29=8


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# 3

x%2B2y-3=0 Start with the given equation.


x%2B2y=3 Add 3 to both sides.


2y=3-x Subtract x from both sides.


2y=-x%2B3 Rearrange the terms.


y=%28-x%2B3%29%2F2 Divide both sides by 2 to isolate y.


y=-%281%2F2%29x%2B3%2F2 Break up the fraction and simplify.


We can see that the equation y=-%281%2F2%29x%2B3%2F2 has a slope m=-1%2F2 and a y-intercept b=3%2F2.


Now to find the slope of the perpendicular line, simply flip the slope m=-1%2F2 to get m=-2%2F1. Now change the sign to get m=2. So the perpendicular slope is m=2.


Now let's use the point slope formula to find the equation of the perpendicular line by plugging in the slope m=-1%2F2 and the coordinates of the given point .


y-y%5B1%5D=m%28x-x%5B1%5D%29 Start with the point slope formula


y--4=2%28x-6%29 Plug in m=2, x%5B1%5D=6, and y%5B1%5D=-4


y%2B4=2%28x-6%29 Rewrite y--4 as y%2B4


y%2B4=2x%2B2%28-6%29 Distribute


y%2B4=2x-12 Multiply


y=2x-12-4 Subtract 4 from both sides.


y=2x-16 Combine like terms.


So the equation of the line perpendicular to x%2B2y-3=0 that goes through the point is y=2x-16.


Here's a graph to visually verify our answer:

Graph of the original equation x%2B2y-3=0 (red) and the perpendicular line y=2x-16 (green) through the point .



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# 5

x%5E2-8x%2By%5E2-6y=0 Start with the given equation


%28x%5E2-8x%29%2B%28y%5E2-6y%29=0 Group like terms.


%28x%5E2-8x%2Bhighlight%2816%29%29%2B%28y%5E2-6y%29=0%2Bhighlight%2816%29 Take half of the x-coefficient -8 to get -4. Square -4 to get 16. Add this value to both sides.


%28x%5E2-8x%2B16%29%2B%28y%5E2-6y%2Bhighlight%289%29%29=0%2B16%2Bhighlight%289%29 Take half of the y-coefficient -6 to get -3. Square -3 to get 9. Add this value to both sides.


%28x-4%29%5E2%2B%28y-3%29%5E2=25 Combine like terms.


%28x-4%29%5E2%2B%28y-3%29%5E2=5%5E2 Rewrite 25 as 5%5E2


Now the equation is in the form %28x-h%29%5E2%2B%28y-k%29%5E2=r%5E2 (which is a circle) where (h,k) is the center and "r" is the radius

In this case, h=4, k=3, and r=5


So the center is (4,3) and the radius is 5 units.


Here's the graph: