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anonymous
 4 years ago
Let f(x) = 3 x^2  8.
Find an equation of the line tangent to the graph of f at the point where x = 3.
y =
anonymous
 4 years ago
Let f(x) = 3 x^2  8. Find an equation of the line tangent to the graph of f at the point where x = 3. y =

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0that's the slope, can you figure out the rest?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0plug 3 into the original equation, to find the corresponding y value, then use the (x,y) point you get with the slope m=18 with the point slope formula yy1=m(xx1) to solve for the line

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0You have to find the point that corresponds to x = 3 on your graph f(x) We get: f(3) = 3(3)^2 8 = 27  8 =19 Then you have to find the derivative of the function, to find the slope of the point at x = 3 Taking the derivative, f'(x) = 3*2x  0 Then, f'(x) = 6x To find the slope at point x= 3, substitute x = 3 in. We get f'(3) = 18 Now we now the point on f(x), which is (3, 19), and the slope at x= 3, which is 18 Now we use the equation of the line formula which is yy0 = m(xx0), Now we just substitute our values in and get: y19 = 18(x3) y19 = 18(x+3)
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