SOLUTION: Please help with this question and provide the workings. Find an equation of the tangent line to the graph of the function f at the point (1, 15). f(x)=〖3x〗^2-11x+23

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Question 992048: Please help with this question and provide the workings. Find an equation of the tangent line to the graph of the function f at the point (1, 15). f(x)=〖3x〗^2-11x+23
Found 2 solutions by stanbon, rothauserc:
Answer by stanbon(75887)   (Show Source): You can put this solution on YOUR website!
Please help with this question and provide the workings.
Find an equation of the tangent line to the graph of the function f at the point (1, 15). f(x)=〖3x〗^2-11x+23
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f'(x) = 6x-11
f'(1) = -5
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Equation of line with m = -5 passing thru (1,15)
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Form:: y = mx + b
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Solve for "b"::
15 = 1*-5 + b
b = 20
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Equation
y = -5x+20
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Cheers,
Stan H.
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Answer by rothauserc(4718)   (Show Source): You can put this solution on YOUR website!
f(x) = 3x^2 -11x + +23
f'(x) is the first derivative of f(x)
f'(x) = 6x -11
f'(x) gives us the slope of the tangent line at any x on f(x)
we are given the point (1,15), therefore
slope of tangent line at x = 1 is f'(1) = 6 - 11 = -5
so far we have
y = -5x + b
but we are given the point (1,15)
15 = -5 + b
b = 20
therefore the equation of the line tangent to f(x) at (1,15) is
y = -5x + 20
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