Constance Said:

Graphing x and y coordinates equation?

We Answered:

It's hard to answer mat questions not knowing exactly what you're studying (in a lot of cases, problems can be solved in more than one way). Here are a few suggestions though. You might start by rearranging the equations (y= everything else) when you can, that is a standard way to graph them. If you are unsure what a graph will look like, plug in a few numbers for x (-2, -1, 0, 1, and 2 are usually pretty accurate), find y and sketch the graph.

Lillian Said:

Math help on graphing quadratics please?

We Answered:

in this question, you should always complete the square.

1)

g(x) = 2x^2 - 6x + 2 - a >0
therefore, the curve has a minimum point.


g(x) = 2x^2 - 6x + 2
g(x) = 2 [ x^2 - 3x + 1 ]
g(x) = 2 [ x^2 - 3x + (3/2)^2 - (3/2)^2 + 1]
g(x) = 2 [ (x + 3/2)^2 - (3/2)^2 + 1]
g(x) = 2 [ (x + 3/2)^2 - 5/4]
g(x) = 2(x + 3/2)^2 - 5/2

the minimum value of the function g(x) is -5/2 when x = -3/2

2)

h(x) = (2 - x) (5 + x) - expand this
h(x) = -2x^2 - 3x + 10

h(x) = -2x^2 - 3x + 10 - a<0
therefore the curve has a maximum point.

now, complete the square

h(x) = -2[x^2 + 3/2x - 5]
h(x) = -2[x^2 + 3/2x + [(3/2)/2]^2 - [(3/2)/2]^2 -5]
h(x) = -2[(x + 3/4)^2 - 9/16 -5]
h(x) = -2[(x + 3/4)^2 - 71/16]
h(x) = -2(x + 3/4)^2 + 71/8

therefore, the maximum value of the function h(x) is 71/8 when x = -3/4
______________________________

not quite sure with this one

edit: I GOT IT!! I REMEMBERED!! OKAY....

1) f(x) = x^2 - 4x - 3

a)the vertex of its graph

If the quadratic is written in the form y = a(x – h)2 + k, then the vertex is the point (h, k).

first, complete the square

f(x) = x^2 - 4x - 3
f(x) = x^2 - 4x + (4/2)^2 - (4/2)^2 -3
f(x) = (x + 2)^2 - 4 -3
f(x) = (x + 2)^2 -7

therefore the vertex of the graph is (-2, -7)

2) h(x) = 2(x - 7)(x + 5)

a) the vertex of its graph

same technique...!

but now, you'll have to expand it first.

h(x) = (2x - 14) (x + 5)
h(x) = 2x^2 + 10x - 14x - 70
h(x) = 2x^2 - 4x - 70
h(x) = 2[ x^2 - 2x - 35]
h(x) = 2[ x^2 - 2x + (2/2)^2 - (2/2)^2 - 35]
h(x) = 2[ (x+1)^2 - 1 - 35
h(x) = 2[ (x+1)^2 - 36]
h(x) = 2(x+1) - 72

therefore the vertex of the graph is ( -1, -72 )

unfortunately, this (question a) is all i remembered.. sorry.


______________________________

general equation for forming a quadratic equation :

x^2 - (sum of roots)x + (product of roots) = 0

Sum of Roots = r1 + r2
Product of Roots = r1r2

hence,

x^2 - ( r1 + r2 ) x + (r1r2) = 0

1) r1 + r2 = 3, r1r2 = 4

x^2 - (3)x + 4 = 0
x^2 - 3x + 4 = 0

2) r1 + r2 = 0, r1r2 = -1

x^2 - 0x + (-1) = 0
x^2 -1 = 0

________________________

Find a quadratic equation with the given roots.

1) 3, -2

x = 3
x = -2

x - 3 = 0
x + 2 = 0

(x - 3) (x + 2) = 0

now, expand

x^2 + 2x - 3x - 6 = 0
x^2 -x -6 = 0

2) 0, 7

x = 0
x = 7

x = 0
x - 7 = 0

x(x - 7) = 0

now, expand

x^2 - 7x = 0

i tried my very best in this... sorry i took so long. it has been a year since i've done this. i love this topic. :D...! hope i helped!!

edit: you're welcome!! :D

Connie Said:

i don't get this math problem on slope-intercept equations?

We Answered:

Ok so for each equation you are going to fill in these values for x on the table:

2,1,0,-1,-2

To find the y values that correspond you simply plug in each value and solve for x. I will show you using the first equation:

y=2x+6
for x=2: y=2*2+6 > y=10
for x=1: y=2*1+6 > y=8
for x=0: y=2*0+6 > y=6
for x=-1: y=2*-1+6 > y=5
for x=-2: y=2*-2+6 > y=2

Once you have both the x and y tables filled in you can start graphing using those coordinates. For example, the first equation you will plot (2, 10) (0, 6) (1, 8) (-1, 5) (-2, 2)

If you get a decimal answer just plug in different x values till you get a whole number! Good luck!

Jean Said:

How would you find an equation on a graph?

We Answered:

easy answer.

work on very bottom.

plot those points on a graph.(optional)
(-6,11)(-3,7)(0,3)(3,-1)(6,-5)(9,-9)
then figure out which one hits the y-axis
(hint: its (0,3))
then figure out the next closest piont to (0,3)
so just subtract (0,3) and another point (try a small one) like (3,-1) and you get -4/3
then you just do y-3= -4/3(x-0)

distribute
y-3= -4/3x

add on both sides

y= -4/3x+3

theres your answer. now heres the work.

3+1/3-0 = -4/3 <--slope
y-3= -4/3(x-0)
y-3= -4/3x
y= -4/3x+3 <--answer

your welcome just come find me if you need any more help.

Jeff Said:

Can you please help me with my math?

We Answered:

You have two variables, x and y. The value for x determines the value for y. Plug in x values to find y values. For instance, for the y = 2x - 1 problem, plug in any x value to find the y value. If x = 0, then y = 2(0) - 1 = -1. If x = 1, then y = 2(1) -1 = 1. If x = 2, then y = 2(2) - 1 = 3 and so forth. After you have these values, you can then create a chart with the x values on the left side and their y values on the right side. The above would look like this:

x y
0 -1
1 1
2 3

Each (x,y) pair is called a coordinate. Coordinates are plotted as points on the graph. You go x values to the right and then y up to plot the point (if one of the values is negative, then you go in the opposite direction). So, for the first one above, you go 0 to the right and 1 down (because it is -1) and then plot to the point. For the second point, go 1 to the right and 1 up and plot the point. For the third point, go to the right 2 and up 3 and plot the point. After you have enough points (2-3 minimum), connect the dots to form the line.

Julian Said:

the point on a graph determined by a coordinate is called a(n)?

We Answered:

The major challenge with math is that it is a totally new, very complex language, and its not treated or taught as such.

First there are variables, where numbers are called letters.
Then theres the rules of the dance, called the identities of algebra.
After that they give you equations, where you relate one variable to another.

Notation of equations is varied.
You may (or may not) have seen something like:

y = 2*x + 1

Its the equation of a straight line. x represents all possible inputs, and for each output there is one and only one output called y. Graphically they form a straight line with a slope of 2 and a y-intercept (the value of y at x=0) of 1.

That expression is equivalent to:
y(x) = 2 * x + 1

The symbol in the brackets, x, is the input. y is the output.

so you have an input that isnt called x, its called n.
And you have an output that isnt called y, its called a.

a(n) isnt a coordinate its a function, unless n isnt a ton of numbers. if n is one and only one number then a(n) should be one and only one point.

I dont know if that helps, but I hope it clears some things up.

Edith Said:

Help on these 8th grade Algebra problems?

We Answered:

Jackson!

Your middle school is not going to be happy when is found out you are not doing your own homework.