Question

how do i start this?
Find an equation for all the lines which pass through the origin and which are tangent to the parabola 푦=−푥^2+6푥−4. There may be 1, 2, or more lines.

Answer 1

whoa i don't know what THAT was but it's supposed to say
\[y=-x^2+6x-4\]

Answer 2

start by saying that if the line is tangent to the graph then the derivative is the slope, so at that the slope between
\[(0,0)\] and
\[(x,-x^2+6x-4)\] must be
\[-2x+6\]

Answer 3

slope between
\[(0,0)\ \text{ and } (x,-x^2+6x-4)\] is
\[\frac{-x^2+6x-4-0}{x-0}=\frac{-x^2+6x-4}{x}\]

Answer 4

set this equal to
\[-2x+6\] and solve for x

Answer 5

i.e solve
\[\frac{-x^2+6x-4}{x}=-2x+6\]
\[-x^2+6x-4=-2x^2+6x\] etc

Answer 6

you have a quadratic equation, you should get two values for x

Answer 7

at most 2 :)

Answer 8

yes, at most!

Answer 9

an in fact exactly 2, one of which is itself exactly 2 if i am not mistaken

Answer 10

can you go on? i'm stuck :/ thank thus far!

Answer 11

\[-x^2+6x-4=-2x^2+6x\]
\[x^2-4=0\]
\[(x+2)(x-2)=0\]
\[x=-2, x=2\]

Answer 12

thanks!