SOLUTION: find the equation of the tangent line to the curve y=9e^-7x at the point (0,9) y(x)= can someone help please thanks

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Question 202793: find the equation of the tangent line to the curve y=9e^-7x at the point (0,9)
y(x)=
can someone help please
thanks
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
The simplest way to solve this is with Calculus. If you are not in a Calculus (or pre Calculus) class, then you might want to repost this so you can get a different solution.

y+=+9e%5E%28-7x%29
%28dy%29%2F%28dx%29+=+%28-7%29%2A9e%5E%28-7x%29+=+-63e%5E%28-7x%29
To find the slope of the tangent at the point (0, 9) we substitute the x-coordinate into dy/dx:
m+=+-63e%5E%28-7%280%29%29+=+-63e%5E0+=+-63%2A1+=+-63
Now we have the slope: -63. And with a point, (0,9) we can return to first-year Algebra to find the equation of the tangent line. Using the Point-slope form: y+-+y%5B1%5D+=+m%28x+-+x%5B1%5D%29 and substituting -63 for m and (0, 9) for +x%5B1%5D+ and +y%5B1%5D+ respectively we get:
y+-+9+=+-63%28x+-+0%29
This may be an acceptable answer. But often equations of lines are expected in slope-intercept form: y = mx + b. So we will transform the above into slope-intercept form. First we simplify:
y+-+9+=+-63%28x%29+=+-63x
Adding 9 to both sides:
y+=+-63x+%2B+9
And now we have the equation of the requested tangent line, in slope-intercept form.