Algebra 2 help, PLEASE!?

Question

we got summer hw for math, i don't know a lot of the stuff in the packet and it's due tomorrow (first day). i had a bad algebra 1 teacher and an even worse geometry teacher so if you could please help me with these problems that'd be amazing

factor: 12x^2 + 101x - 35

solve: -24x^2 - 78x^2 - 18x = 0

how to graph x= -2 and y=3 on the same coordinate plane

write an equation through the points (1,4) and (-1,3)

find x: x^2 + 6x + 9 + x^2 = 400

explain how to choose the appropriate scales for the x and y axis when graphing y=0.5x^2 - 12

any answers are great, i tried working with a friend and these are the answers we couldn't get. please don't call me lazy, i tried my hardest

Answer 1

Factor:

This one was tricky.  Basically, you multiply the coefficient of x^2 by the final number (12)*(-35) and come up with -420.  Then go through all the factors of -420 until you find a pair that will add up to 101.

The factors are:

1,-420

-1,420

2,-210

-2,210

3,-140

-3,140

4,-105

-4,105 **** ding ding ding ding.

Set up two sets of parentheses.  Fill in factors of 12 in the first slot and factors of 35 in the second slot so the inner terms multiply out to be 105 and the outer terms multiply out to be -4.

(4x+35)(3x-1)

Solve:

Combine like terms:

-102x^2-18x=0

Factor out an x from each of the terms on the left:

x*(-102x-18)=0

Either x=0 or -102x-18=0

-102x-18=0

Add 18 to both sides

-102x=18

Divide both sides by -102

x=-18/102=-3/17

How to graph:  Draw your coordinate plane.  x=-2 is a vertical line through (-2,0).  Draw that in.  y=3 is a horizontal line through (0,3). Draw that in.

Write an equation.

First, find the slope.  The slope is the difference of the y's divided by the difference of the x's

Difference of the y's: 4-3=1

Difference of the x's: 1-(-1)=1+1=2

Slope= 1/2

Now, the easiest way to write the equation of a line if you have a point and a slope is to use the point-slope form (y-y1)=m(x-x1)

(y-4)=1/2(x-1)

Find x.  First, combine like terms

2x^2+6x+9=400

Subtract 400 from both sides

2x^2+6x-391=0

Use quadratic formula to solve from there with a=2, b=6, and c=-391

How to choose the appropriate scales for x and y axis.  I think this has to do with the shape of the graph.  Notice that this is a parabola (because it has the form y=x^2) that opens upward (because the coefficient of x^2 is positive) and it's shifted down 12 units (because of the -12 at the end).  The lowest number you're going to have on your graph is -12, so make your scale small enough that you can start at -12 and sketch a parabola that opens upward.  Hope that helps.

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