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

Question

anonymous · Answer

JIm you are seriously my hero

jim_thompson5910 · Answer

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.

jim_thompson5910 · Answer

oops made a typo, one sec

jim_thompson5910 · Answer

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)

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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 

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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.

jim_thompson5910 · Answer

ok, fixed the typo

anonymous · Answer

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

anonymous · Answer

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

jim_thompson5910 · Answer

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