1. ## Just need some help please

Hi for class i have a hundred problem assignment and out of all the problems a few have stumped me i know there not difficult but i am having trouble, i just want to do really good on this assignment , if you can help thanks.

1.)The business opened with a debt of $4300. After 2 years, it accumulated profit of$4600. Find the profit as a function of time , knowing the profit function is linear.

2.)) Find the slope and the y-intercept of the line

3.)The equation of the line with slope that goes through the point can be written in the form where is

4.)Find the - and -intercepts of the graph of the equation

again thank you so uch of you can help

2. heres #2

$y=10(9x - 9)$

first change it so it looks like $y=mx+b$

$y=90x-90$ so $m=90$

to find the y intercept set $x=0$

$y=-90$

3. Hello Duffmac

Welcome to Math Help Forum!
Originally Posted by Duffmac
Hi for class i have a hundred problem assignment and out of all the problems a few have stumped me i know there not difficult but i am having trouble, i just want to do really good on this assignment , if you can help thanks.

1.)The business opened with a debt of $4300. After 2 years, it accumulated profit of$4600. Find the profit as a function of time , knowing the profit function is linear.
I'm not exactly sure what the question means by the word 'accumulated'. Does it mean (a) that in two years it made a profit of $4600, and so was now$300 in surplus? Or does it mean (b) that in two years it made enough profit to wipe out the debt and is now $4600 in surplus? (It would need to have made$8900 profit to do this.) Here are both answers.

(a) If the function is linear, then it means that the profit each year is constant. So that's $4600\div2 = 2300$ per year. After t years, then, the profit will be $P(t)=2300t$.

(b) In a similar way, this time
$P(t) = -4300 + 4450t$.
2.)) Find the slope and the y-intercept of the line
$y = 10(9x-9)$
$=90x -90$
This is now in the form $y = mx +c$, so the slope is $90$, and the intercept is $-90$.

3.)The equation of the line with slope that goes through the point can be written in the form where is
The value of $m$ is $-1$ (the slope). If you want the value of $b$ as well, plug in the values of $x$ and $y$ that you know (and this value of $m$), to get:
$-2=(-1)(5)+b$
Can you solve this equation now to find $b$?

4.)Find the - and -intercepts of the graph of the equation
The $y$-intercept is where $x$ has the value $0$. So that's $y =-40$.

For the $x$-intercepts (there are two), give $y$ the value $0$:
$x^2+3x-40=0$

$\Rightarrow (x+8)(x-5)=0$
Can you complete this quadratic equation?

Hello Duffmac

Welcome to Math Help Forum! I'm not exactly sure what the question means by the word 'accumulated'. Does it mean (a) that in two years it made a profit of $4600, and so was now$300 in surplus? Or does it mean (b) that in two years it made enough profit to wipe out the debt and is now $4600 in surplus? (It would need to have made$8900 profit to do this.) Here are both answers.

(a) If the function is linear, then it means that the profit each year is constant. So that's $4600\div2 = 2300$ per year. After t years, then, the profit will be $P(t)=2300t$.

(b) In a similar way, this time $P(t) = -4300 + 4450t$.
$y = 10(9x-9)$
$=90x -90$
This is now in the form $y = mx +c$, so the slope is $90$, and the intercept is $-90$.

The value of $m$ is $-1$ (the slope). If you want the value of $b$ as well, plug in the values of $x$ and $y$ that you know (and this value of $m$), to get:
$-2=(-1)(5)+b$
Can you solve this equation now to find $b$?

The $y$-intercept is where $x$ has the value $0$. So that's $y =-40$.

For the $x$-intercepts (there are two), give $y$ the value $0$:
$x^2+3x-40=0$

$\Rightarrow (x+8)(x-5)=0$
Can you complete this quadratic equation?