Find Equation Of Line Tangent To Graph Of F At X=-1 | Step-By-Step Guide

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 the equation of the line tangent to the graph of f at x = -1, we need to use the point-slope form of the equation of a line, which is:

y – y1 = m(x – x1)

where y1 and x1 are the coordinates of the given point on the line (in this case, (-1, f(-1))), and m is the slope of the tangent line (in this case, f'(-1)).

So, first, we need to find f'(-1):

f'(x) = -3x + 4

f'(-1) = -3(-1) + 4 = 7

So, the slope of the tangent line is 7. Now, we need to find the y-coordinate of the point on the line by plugging in x = -1:

f(-1) = 6

So, the point on the line is (-1,6).

Plugging these values into the equation of the line, we get:

y – 6 = 7(x + 1)

Simplifying and putting in slope-intercept form, we get:

y = 7x + 13

Therefore, the equation of the line tangent to the graph of f at x = -1 is y = 7x + 13.

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