Question

Help me with this problem in the picture?

Answer 1

Ignore the bold text at the bottom.

Answer 2

if you take a peek at the graph... where is the graph TOUCHING the x-axis?
where is it TOUCHING the y-axis?

Answer 3

it's touching the y at 3 and x at -4

Answer 4

so... when it touches the x-axis, is -4, what's "y"? well y = 0 at that point, so the point there will be (-4, 0)

it touches the y-axis at 3, what's "x"? well, x =0 at that point, so the point for that will be (0, 3)

so we have 2 points, what's our slope?

well, recall that \(\bf slope = m= \cfrac{rise}{run} \implies \cfrac{y_2-y_1}{x_2-x_1}\\ \quad \\
(-4,0)\qquad (0,3)\implies \cfrac{3-0}{0-(-4)}\implies\cfrac{3}{4}\)

Answer 5

THank you very much!

Answer 6

so now we know the slope and 2 points
so let us use the point-slope form of \(\large y-y_1=m(x-x_1)\)

m = slope

and plug in the known values, say let's use either point... say (3, 0)

\(\bf (3, 0)\quad m=\cfrac{3}{4}\implies y-y_1=m(x-x_1)\implies y-(3)=\cfrac{3}{4}(x-0)\)

solve for "y" to get the equation for the line

Answer 7

hmm I got my values switched there one sec

Answer 8

Alrigty

Answer 9

\(\bf (3, 0)\quad m=\cfrac{3}{4}\implies y-y_1=m(x-x_1)\implies y-(3)=\cfrac{0}{4}(x-3)\)

Answer 10

... what the ? shoot.... lemme ... redo that

Answer 11

\(\bf (3, 0)\quad m=\cfrac{3}{4}\implies y-y_1=m(x-x_1)\implies y-(0)=\cfrac{3}{4}(x-3)\)

Answer 12

so if you solve for "y", that'd get the equation of the line

Answer 13

Ahh okay. Thanks a bunch. :)

Answer 14

yw