1. ## Graph Help

a)x2 + 4x + 3 =

b) x2 - 5x + 6

c) x2 + 2x – 3

d)x2 – 2x – 8

2)Give the values of m and c

a)y = 2x + 3

b) y = 4x – 2

c)y = 2x – 5

d)y = 1/2x – 8

e) 3y = 6x – 9

f) 1/2y = 2x + 3

These are the answers I came up for question no #2. Comparing the given equations with the standard format of
y = mx + c

------------------------------------------------------------
a)
y = 2x + 3

m = 2 and c = 3

--------------------------------------------------------------
b)
y = 4x - 2

m = 4 and c = -2

---------------------------------------------------------------
c)
y = 2x - 5

m = 2 and c = -5

--------------------------------------------------------------
d)
y = 1/2x - 8

m = 1/2 and c = -8

---------------------------------------------------------------
e)
3y = 6x - 9

rearranging y = 2x -3

m = 2 and c = -3

---------------------------------------------------------------
f)
1/2y = 2x + 3

rearranging y = 4x + 6

m = 4 and c = 6

-----------------------------------------------------------------

2. Originally Posted by Soul

a)x2 + 4x + 3

b) x2 - 5x + 6

c) x2 + 2x – 3

d)x2 – 2x – 8

2)Give the values of m and c

a)y = 2x + 3

b) y = 4x – 2

c)y = 2x – 5

d)y = 1/2x – 8

e) 3y = 6x – 9

f) 1/2y = 2x + 3
By "solve" the first set do you mean FACTOR?

And I'm slightly confused on how you have your examples set up. Is it set up the way I have you quoted?

3. Originally Posted by Soul

a)x2 + 4x + 3
$x^2+4x+3=0$ It's in this form $ax^2+bx+c=0$ in your problem it's a=1 , b=4 , c=3.
First you find $b^2-4*a*c$ which here is $D = 4^2-4*1*3=16-12=4$ and then you use $\frac{-b+\sqrt{D}}{2a}$ to find the first solution and $\frac{-b-\sqrt{D}}{2a}$ to find the second.
Here the solutions are : $\frac{-4+2}{2}=-1$ and $\frac{-4-2}{2}=-3$

Try the rest...

4. Originally Posted by Soul
2)Give the values of m and c

a)y = 2x + 3b) y = 4x – 2c)y = 2x – 5

a)y = 1/2x – 8e) 3y = 6x – 9f) 1/2y = 2x + 3
What do you mean to give the values of m and c ???
What is m and c ?

5. Originally Posted by SENTINEL4
What do you mean to give the values of m and c ???
What is m and c ?
I suspect the OP means to compare them to the formula for a straight line - often denoted as $y = mx + c$

In relation to post 3 if the discriminant is a perfect square that means there are rational roots and you should be able to factor that equation. For example $x^2+4x+3 = (x+3)(x+1)$

6. Originally Posted by Soul

a)x2 + 4x + 3
b) x2 - 5x + 6
c) x2 + 2x – 3
b)x2 – 2x – 8
These are expressions; having no "equals" signs in them, they cannot be "equations". Please reply with clarification.

Originally Posted by Soul
2)Give the values of m and c

a)y = 2x + 3
b) y = 4x – 2
c)y = 2x – 5
a)y = 1/2x – 8
e) 3y = 6x – 9
f) 1/2y = 2x + 3
For the first four, the equations are already written in the form "y = mx + c". Copy off the values of "m" and "c".

For the last two, solve for "y=", and then copy off the values.

7. Oh that makes sense. Well if these equations are set to zero than an easier method (if you are comfortable factoring) is to do so:

$x^2+4x+3$ factors to $(x+3)(x+1)$. And thus the resultant equation:

$(x+3)(x+1)=0$ we can solve by taking any of the two terms and setting it individually equal to zero:

$(x+3)=0\\(x+1)=0$

Using the quadratic equation also works, but might be a bit of a waste on equations that factor nicely.

On the second set, "Give the values of M and C" I'm assuming refers to:

$y=mx+b$;

where $M$ is the slope of our equation and $b$ is our y-intercept and the equation is the slope-intercept form of a line.

Thusly, arranging each of the equations a-f and putting them in slope-intercept form will yield our M and C.