Find the equation of the tangent line to the graph of the function at the given value of x. f(x) =x^2+5x at x=4 please I need step by step?
回答 (2)
2016-02-19 9:54 pm
✔ 最佳答案
Well,f(x) = x^2 + 5x
x = 4 ==> f(4) = 16+20 = 36
f '(x) = 2x + 5
therefore, f '(a) is the slope of the tangent :
f '(4) = 2*4 + 5 = 13
therefore, the equation at A(4, f(4) ) is :
y - f(4) = f '(4) (x - 4)
y - 36 = 13x - 52
y = 13x - 16 <--- answer
hope it' ll help !!
2016-02-19 10:00 pm
y = f(x) = x² + 5x
When x = 4 : y = 36
Thus, (4, 36) is a point lying on the given graph.
f'(x) = 2x + 5
Slope of the tangent line at (x = 4)
= f'(4)
= 13
The tangent line has a slope 13 and passes through (4, 36).
Thus, the equation of the tangent line is :
(y - 36)/(x - 4) = 13
13(x - 4) = y - 36
13x - 52 = y - 36
13x - y - 16 = 0
When x = 4 : y = 36
Thus, (4, 36) is a point lying on the given graph.
f'(x) = 2x + 5
Slope of the tangent line at (x = 4)
= f'(4)
= 13
The tangent line has a slope 13 and passes through (4, 36).
Thus, the equation of the tangent line is :
(y - 36)/(x - 4) = 13
13(x - 4) = y - 36
13x - 52 = y - 36
13x - y - 16 = 0
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