How do I graph x-4y=4 and 3x+y=-1

Mathematics

1 answer:

ololo11 [35]3 years ago

8 0

Answer:

Draw all these points and draw a straight line through all of them.

Step-by-step explanation:

y = mx + b

x - 4y = 4

-x -x

-4y = -x + 4

/-4 /-4

y = $\frac{1}{4}x$ -1

(0,-1),(4,0),(-8,-3),(-4,-2),(8,1) - The first graph attached

3x+y=-1

y = -3x - 1

(0,-1),(-1,2),(-3,8),(2,-7),(3,-10) - The second/last graph attached.

Hope this helps!

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Please help thank you

Dvinal [7]

the tale-tell fellow is the number inside the parentheses.

if that number, the so-called "growth or decay factor", is less than 1, then is Decay, if it's more than 1, is Growth.

$\bf f(x)=0.001(1.77)^x\qquad \leftarrow \qquad \textit{1.77 is greater than 1, Growth} \\\\[-0.35em] ~\dotfill\\\\ f(x)=2(1.5)^{\frac{x}{2}}\qquad \leftarrow \qquad \textit{1.5 is greater than 1, Growth} \\\\[-0.35em] ~\dotfill\\\\ f(x)=5(0.5)^{-x}\implies f(x)=5\left( \cfrac{05}{10} \right)^{-x}\implies f(x)=5\left( \cfrac{1}{2} \right)^{-x} \\\\\\ f(x)=5\left( \cfrac{2}{1} \right)^{x}\implies f(x)=5(2)^x\qquad \leftarrow \qquad \textit{Growth} \\\\[-0.35em] ~\dotfill$

$\bf f(t)=5e^{-t}\implies f(t)=5\left( \cfrac{e}{1} \right)^{-t}\implies f(t)=5\left( \cfrac{1}{e} \right)^t \\\\\\ \cfrac{1}{e}\qquad \leftarrow \qquad \textit{that's a fraction less than 1, Decay}$

now, let's take a peek at the second set.

$\bf f(x)=3(1.7)^{x-2}\qquad \leftarrow \qquad \begin{array}{llll} \textit{the x-2 is simply a horizontal shift}\\\\ \textit{1.7 is more than 1, Growth} \end{array} \\\\[-0.35em] ~\dotfill\\\\ f(x)=3(1.7)^{-2x}\implies f(x)=3\left(\cfrac{17}{10}\right)^{-2x}\implies f(x)=3\left(\cfrac{10}{17}\right)^{2x} \\\\\\ \textit{that fraction is less than 1, Decay} \\\\[-0.35em] ~\dotfill$

$\bf f(x)=3^5\left( \cfrac{1}{3} \right)^x\qquad \leftarrow \qquad \textit{that fraction is less than 1, Decay} \\\\[-0.35em] ~\dotfill\\\\ f(x)=3^5(2)^{-x}\implies f(x)=3^5\left( \cfrac{2}{1} \right)^{-x}\implies f(x)=3^5\left( \cfrac{1}{2} \right)^x \\\\\\ \textit{that fraction in the parentheses is less than 1, Decay}$

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