find the equation … - QuestionCove
OpenStudy (anonymous):

find the equation of a line passing through the vertex of y=2(x-3)^2+5 and the origin

6 years ago
OpenStudy (anonymous):

JIm you are seriously my hero

6 years ago
OpenStudy (jim_thompson5910):

So it's not too bad, you just need to know how to find the vertex and how to find the equation of a line through two points.

6 years ago
OpenStudy (jim_thompson5910):

oops made a typo, one sec

6 years ago
OpenStudy (jim_thompson5910):

Vertex of y=a(x-h)^2+k is (h,k). In this case, h=3 and k=5. So the vertex is (3,5) The origin is a special point located at the center of the axis and it is the point (0,0) ----------------------------------------------------- Now we must find the slope of the line going through the points (3,5) and (0,0) m = (y2-y1)/(x2-x1) m = (0-5)/(0-3) m = (-5)/(-3) m = 5/3 ------------------------------------------------------- Now use the general equation y = mx+b, plug in m=5/3, x=0 and y=0 (one of the points) to get y=mx+b 0=(5/3)(0)+b Now solve for b: 0=(5/3)(0)+b 0=0+b 0=b b=0 So plug this last piece of info into y = mx+b to get y = (5/3)x+0 which simply means the equation is $y=\frac{5}{3}x$ which is the answer.

6 years ago
OpenStudy (jim_thompson5910):

ok, fixed the typo

6 years ago
OpenStudy (anonymous):

lmao. Cool give me a second. Got another great question to do with elipse in a second

6 years ago
OpenStudy (anonymous):

the vertext isn't 3 and -5???

6 years ago
OpenStudy (jim_thompson5910):

No, the vertex is (3,5). I made the typo in saying that it was (-3,5), but I fixed that.

6 years ago