The derivative of the function f is given by f′(x)=−3x+4 for all x, and f(−1)=6. Which of the following is an equation of the line tangent to the graph of f at x=−1 ?
y=7x+13The slope of the line tangent to the graph of ff at x=−1 is f′(−1)=−3(−1)+4=7. An equation of the line containing the point (−1,6) with slope 7 is y=7(x+1)+6=7x+13
To find an equation of the line tangent to the graph of f at x=-1, we need to use the point-slope form of a linear equation:
y – y1 = m(x – x1)
where (x1, y1) is the point of tangency and m is the slope of the tangent line.
We know that the point of tangency is (-1, f(-1)) = (-1, 6). To find the slope of the tangent line at x=-1, we simply plug in x=-1 into f'(x):
f'(-1)= -3(-1) + 4 = 7
Therefore, the slope of the line tangent to the graph of f at x=-1 is 7.
Now we can plug in the values we have into the point-slope form:
y – 6 = 7(x + 1)
Simplifying, we get:
y = 7x + 13
So an equation of the line tangent to the graph of f at x=-1 is y = 7x + 13.
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