Find an equation of the line tangent to the graph of
y=1−8x−9x^2
at the point (3, -104).

Question

Find an equation of the line tangent to the graph of
y=1−8x−9x^2
at the point (3, -104).

anonymous · Answer

to find a tangent line, take the derivative of the function, which will give you the slope at any point. in this case the derivative is -8-18x. then plug in the x value of the original equation to find the slope at that point: -8-18(3)=-62

anonymous · Answer

then use the point slope formula y-y1 = m(x-x1) and plug in your point and the slope just given. theres nothing to solve for (no value for x and y) in the point slope form, just an equation of form y=mx+b

anonymous · Answer

y+104=-62(x-3)

anonymous · Answer

yes. then you can multiply out and simplify a little if you want to get a form y=mx+b, but thats the right equation

anonymous · Answer

y=-62x+82

anonymous · Answer

@etemplin thats not right

anonymous · Answer

ok hold on...let me see what i did wrong. Im sorry

anonymous · Answer

its all good

anonymous · Answer

logically it appears to be right because 1) the functions have the same value at x=3 (the value is -104), and 2) the slope of the function 1-8x-9x^2 is given by the function -8-18x, evaluated at 3 gives -62 as the slope of the graph and the tangent line. I have checked by using the calculator and wolframalpha.com. is it possible theres a typo in your original data?

anonymous · Answer

or possibly a formatting issue (since everything is in one line, there may be a misconception about the exponent, or possibly a negative sign missed, etc)

anonymous · Answer

ok  my bad it was my bad entered it in wrong.