Question

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? I think it’s y=7x+13, but I could just be delusional

Answer 1

Answer:

You are correct. The answer is $y=7x+13$

Step-by-step explanation:

First, we need to plug in our x value to the derviative's equation to find the slope of the line tangent to the graph, f

$f'(x)=-3x+4\\\\f'(-1)=-3(-1)+4\\\\f'(-1)=3+4\\\\f'(-1)=7$

Now that we have found the slope and have a given point: (-1,6) we can use the point slope formula to find the equation of the line tangent to the graph f.

$y-6=7(x+1)\\\\y-6=7x+7\\\\y=7x+13$

Answer 2

The required equation is 7x + 13

Given the derivative of the function, f is given by f′(x)=−3x+4

Get the gradient by integrating the given function as shown:

f′(x)=−3x+4

Substitute x = -1 into the function to get the slope

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

f'(-1) = 3 + 4

f'(-1) = 7

Hence the slope is 7

Get the required equation in point-slope form.

Recall that $y-y_0 = m(x-x_0)$

Given that;

• m = 7
• $(x_0,y_0) =(-1, 6)$

Substitute into the expression;

$y-6 = 7(x-(-1)\\y-6=7(x+1)\\y-6=7x+7\\y=7x+7+6\\y=7x+13$

Therefore the required equation is 7x + 13

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