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Mathematics 15 Online
OpenStudy (heart_offire):

Find the equation of the line. A) y=- 3 - 2 x + 1 B) y=- 2 - 3 x - 1 C) y= 2 - 3 x + 1 D) y= 3 _ 2 x - 1 http://www.usatestprep.com/modules/gallery/files/4/451/451.png i took out b and c

OpenStudy (jdoe0001):

hmmm when y = + 5 on that graphic, what is the value of x?

OpenStudy (heart_offire):

@jdoe0001 -1

OpenStudy (jdoe0001):

check closely, notice the graph|dw:1453421732244:dw|

OpenStudy (heart_offire):

@jdoe0001 so 0

OpenStudy (jdoe0001):

notice where y is 5|dw:1453422053982:dw|

OpenStudy (heart_offire):

oh ok i thought you mean on the x axis where it intercepts

OpenStudy (heart_offire):

so the x value is 5

OpenStudy (jdoe0001):

hmmm is it? check closely where the line on the grid hits the x-axis

OpenStudy (heart_offire):

on the graph that i gave the link for it lines up perfectly with 5

OpenStudy (jdoe0001):

hmmm it lines up with 5 on the y-axis, yes, but it hits the x-axis elsewhere

OpenStudy (heart_offire):

now im confused because the x axis has two side, negative and positive. on the negative side the lines goes through between -1 and 0

OpenStudy (jdoe0001):

actually, it doesn't touch -1 on the x-axis, when the y-axis is 0 more like -1.5 or -1,5,0 but is not exact, reason why I asked on the y = +5 instead since when y = +5 and -5, there's a defined x-axis value, check closely

OpenStudy (heart_offire):

6

OpenStudy (jdoe0001):

yeap, is x = 6 when y = +5 what about when y = -5?

OpenStudy (heart_offire):

-9

OpenStudy (jdoe0001):

yeap so....now let us take a peek at those 2 points 6, 5 and -9, -5 we'll need to get the slope first one sec

OpenStudy (jdoe0001):

\(\bf \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) &({\color{red}{ 6}}\quad ,&{\color{blue}{ 5}})\quad % (c,d) &({\color{red}{ -9}}\quad ,&{\color{blue}{ -5}}) \end{array} \\\quad \\ % slope = m slope = {\color{green}{ m}}= \cfrac{rise}{run} \implies \cfrac{{\color{blue}{ -5}}-{\color{blue}{ 5}}}{{\color{red}{ -9}}-{\color{red}{ 6}}}\implies \cfrac{\cancel{-10}}{\cancel{-15}}\implies \cfrac{2}{3} \\ \quad \\ % point-slope intercept y-{\color{blue}{ 5}}={\color{green}{ \cfrac{2}{3}}}(x-{\color{red}{ 6}})\qquad \textit{plug in the values and solve for "y"}\\ \qquad \uparrow\\ \textit{point-slope form}\)

OpenStudy (heart_offire):

so if we change that to y=mx+b the answer would be C

OpenStudy (jdoe0001):

well, dunno what C is, but let us see what it's by solving for "y" :) \(\bf y-{\color{blue}{ 5}}={\color{green}{ \cfrac{2}{3}}}(x-{\color{red}{ 6}})\implies y=\cfrac{2}{3}x-\cfrac{2}{\cancel{3}}\cdot \cancel{6}+5 \\ \quad \\ y=\cfrac{2}{3}x-4+5\implies y=\cfrac{2}{3}x+1\)

OpenStudy (heart_offire):

thank you

OpenStudy (jdoe0001):

yw

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