Question 202793
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.<br>
{{{y = 9e^(-7x)}}}
{{{(dy)/(dx) = (-7)*9e^(-7x) = -63e^(-7x)}}}
To find the slope of the tangent at the point (0, 9) we substitute the x-coordinate into dy/dx:
{{{m = -63e^(-7(0)) = -63e^0 = -63*1 = -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[1] = m(x - x[1])}}} and substituting -63 for m and (0, 9) for {{{ x[1] }}} and {{{ y[1] }}} respectively we get:
{{{y - 9 = -63(x - 0)}}}
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(x) = -63x}}}
Adding 9 to both sides:
{{{y = -63x + 9}}}
And now we have the equation of the requested tangent line, in slope-intercept form.