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3 years ago

How do i graph y=-3/4x-1

Mathematics

2 answers:

Tasya [4]3 years ago

6 0

Answer:

Start at -1 and go down 3 right 4,repeat. Then up 3 left 4 from -1, repeat.

Step-by-step explanation:

morpeh [17]3 years ago

4 0

Answer:

So start at the y intercept which is -1 in this case becuase the point slope formula goes like

y=mx+b

m= slope

b= y-intercept

Y= Ranodm Y coordinate

X= Random X coordinate

You don’t have to worry so much about the x and Y, the main focus while graphing is the slope (m) and the y-intercept (y)

So -1 is the y-intercept so you go to -1 on the graph on the Y axis, it should be right below 0.

Now to make the points, the slope is -3/4

So a slope is written rise/run

Rise is going up and run is going right

-3 is the rise

4 is the run

Since the slope is negative (due to having a negative on 1 side)

We just do the opposite for the rise

Go down 3 and go to the right 4.

There you have it

Also for image reference

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which has partial derivatives (set equal to 0) satisfying

$\begin{cases}L_x=2x+\lambda_1+\lambda_2=0\\L_y=2y+2\lambda_1-\lambda_2=0\\L_z=2z+\lambda_1=0\\L_{\lambda_1}=x+2y+z-4=0\\L_{\lambda_2}=x-y-8=0\end{cases}$

This is a fairly standard linear system. Solving yields Lagrange multipliers of $\lambda_1=-\dfrac{32}{11}$ and $\lambda_2=-\dfrac{104}{11}$ , and at the same time we find only one critical point at $(x,y,z)=\left(\dfrac{68}{11},-\dfrac{20}{11},\dfrac{16}{11}\right)$ .

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$\mathbf H(x,y,z)=\begin{bmatrix}f_{xx}&f_{xy}&f_{xz}\\f_{yx}&f_{yy}&f_{yz}\\f_{zx}&f_{zy}&f_{zz}\end{bmatrix}=\begin{bmatrix}=\begin{bmatrix}2&0&0\\0&2&0\\0&0&2\end{bmatrix}$

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