Question 204503
# 1

{{{f(x+h)=(x+h)^2+3}}} Replace each "x" with "x+h"

{{{f(x+h)=x^2+2xh+h^2+3}}} FOIL (ie expand)

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{{{(f(x+h)-f(x))/h}}} Move onto the given difference quotient.

{{{(x^2+2xh+h^2+3-(x^2+3))/h}}} Plug in {{{f(x+h)=x^2+2xh+h^2+3}}} and {{{f(x)=x^2+3}}}

{{{(x^2+2xh+h^2+3-x^2-3)/h}}} Distribute

{{{(2xh+h^2)/h}}} Combine like terms.

{{{(h(2x+h))/h}}} Factor out the GCF "h" from the numerator.

{{{(cross(h)(2x+h))/cross(h)}}} Cancel out the common terms.

{{{2x+h}}} Simplify

So {{{(f(x+h)-f(x))/h=2x+h}}} when {{{f(x)=x^2+3}}}

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# 2

Take note that when {{{x=-3}}} it is less than zero. Because {{{f(x)=-x}}} when {{{x<0}}}, this means that we simply plug in {{{x=-3}}} to get: {{{f(-3)=-(-3)=3}}}.

So {{{f(-3)=3}}}

Also, when {{{x=2}}} it is greater than zero. Since {{{f(x)=x^3}}} when {{{x>=0}}}, we just plug in {{{x=2}}} to get: {{{f(2)=(2)^3=8}}}.

So {{{f(2)=8}}}

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# 3

{{{x+2y=3}}} Add 3 to both sides.

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

{{{2y=-x+3}}} Rearrange the terms.

{{{y=(-x+3)/2}}} Divide both sides by {{{2}}} to isolate y.

{{{y=-(1/2)x+3/2}}} Break up the fraction and simplify.

We can see that the equation {{{y=-(1/2)x+3/2}}} has a slope {{{m=-1/2}}} and a y-intercept {{{b=3/2}}}.

Now to find the slope of the perpendicular line, simply flip the slope {{{m=-1/2}}} to get {{{m=-2/1}}}. 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=-1/2}}} and the coordinates of the given point *[Tex \LARGE \left$$6,-4\right$$].

{{{y--4=2(x-6)}}} Plug in {{{m=2}}}, {{{x[1]=6}}}, and {{{y[1]=-4}}}

{{{y+4=2(x-6)}}} Rewrite {{{y--4}}} as {{{y+4}}}

{{{y+4=2x+2(-6)}}} Distribute

{{{y+4=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+2y-3=0}}} that goes through the point *[Tex \LARGE \left$$6,-4\right$$] is {{{y=2x-16}}}.

Here's a graph to visually verify our answer:

{{{drawing(500, 500, -10, 10, -10, 10,
graph(500, 500, -10, 10, -10, 10,-(1/2)x+3/2,2x-16)
circle(6,-4,0.08),
circle(6,-4,0.10),
circle(6,-4,0.12))}}}

Graph of the original equation {{{x+2y-3=0}}} (red) and the perpendicular line {{{y=2x-16}}} (green) through the point *[Tex \LARGE \left$$6,-4\right$$].

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# 5

{{{(x^2-8x)+(y^2-6y)=0}}} Group like terms.

{{{(x^2-8x+highlight(16))+(y^2-6y)=0+highlight(16)}}} Take half of the x-coefficient -8 to get -4. Square -4 to get 16. Add this value to both sides.

{{{(x^2-8x+16)+(y^2-6y+highlight(9))=0+16+highlight(9)}}} Take half of the y-coefficient -6 to get -3. Square -3 to get 9. Add this value to both sides.

{{{(x-4)^2+(y-3)^2=25}}} Combine like terms.

{{{(x-4)^2+(y-3)^2=5^2}}} Rewrite 25 as {{{5^2}}}

Now the equation is in the form {{{(x-h)^2+(y-k)^2=r^2}}} (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:

{{{drawing(500, 500, -10, 10, -10, 10,
grid(1),
graph(500, 500, -10, 10, -10, 10,0),
circle(4,3,5)
)}}}