What is the equation of the line that is parallel to the tangent line of f(x)=−x3+7x2−13 at f(−2) assuming the parallel line passes through the point (−4,5) ? (1 point) Responses
First, let's find the slope of the tangent line at f(-2).
To find the slope of the tangent line, we need to find the derivative of f(x):
f(x) = -x^3 + 7x^2 - 13
f'(x) = -3x^2 + 14x
Now let's find the slope at x = -2:
f'(-2) = -3(-2)^2 + 14(-2) = -3(4) - 28 = -12 - 28 = -40
Therefore, the slope of the tangent line at f(-2) is -40.
Since the parallel line has the same slope, the slope of the parallel line is also -40.
Now let's use the point-slope form of a linear equation to find the equation of the parallel line.
The point-slope form is given by: y - y1 = m(x - x1), where (x1, y1) is a point on the line and m is the slope.
We have the point (-4, 5) on the parallel line and the slope (-40).
Using these values in the point-slope form, we get:
y - 5 = -40(x - (-4))
y - 5 = -40(x + 4)
y - 5 = -40x - 160
y = -40x - 155
Therefore, the equation of the line that is parallel to the tangent line of f(x) = -x^3 + 7x^2 - 13 at f(-2) and passes through the point (-4, 5) is y = -40x - 155.
To find the equation of the line that is parallel to the tangent line of the given function at f(-2), we need to find two things: the slope of the tangent line and the point at which the parallel line passes through.
First, let's find the slope of the tangent line at f(-2). To find the slope, we need to take the derivative of the function f(x) and then evaluate it at x = -2.
Taking the derivative of f(x), we have:
f'(x) = -3x^2 + 14x
Now, let's evaluate f'(-2):
f'(-2) = -3(-2)^2 + 14(-2)
= -3(4) - 28
= -12 - 28
= -40
So, the slope of the tangent line at f(-2) is -40.
Now, let's find the equation of the line that passes through the point (-4, 5) with a slope of -40.
We can use the point-slope form of a line to find the equation. The point-slope form is:
y - y1 = m(x - x1)
where (x1, y1) is a point on the line, and m is the slope of the line.
Plugging in the values, we have:
y - 5 = -40(x - (-4))
y - 5 = -40(x + 4)
y - 5 = -40x - 160
Now, let's rearrange the equation to the standard slope-intercept form, y = mx + b:
y = -40x - 160 + 5
y = -40x - 155
Therefore, the equation of the line that is parallel to the tangent line of f(x) = -x^3 + 7x^2 - 13 at f(-2) and passes through the point (-4, 5) is y = -40x - 155.
To find the equation of a line that is parallel to the tangent line of f(x) = -x^3 + 7x^2 - 13 at f(-2), we need to find the slope of the tangent line at f(-2), and then use that slope to find the equation of the parallel line.
Step 1: Find the slope of the tangent line at f(-2)
To find the slope of the tangent line, we need to take the derivative of the function f(x). The derivative of f(x) = -x^3 + 7x^2 - 13 is:
f'(x) = -3x^2 + 14x
Now, we can find the slope by plugging in the x-value of f(-2) into the derivative:
slope = f'(-2) = -3(-2)^2 + 14(-2)
= -3(4) - 28
= -12 - 28
= -40
Step 2: Use the slope to find the equation of the parallel line
We have the slope of -40, and we also have a point on the parallel line, which is (-4, 5). We can use the point-slope form of a line to find the equation of the parallel line:
y - y1 = m(x - x1)
Using the point (-4, 5) and the slope -40, we can plug in these values:
y - 5 = -40(x - (-4))
y - 5 = -40(x + 4)
Now, we can simplify this equation to standard form:
y - 5 = -40x - 160
y = -40x - 155
Therefore, the equation of the line that is parallel to the tangent line of f(x) = -x^3 + 7x^2 - 13 at f(-2), and passes through the point (-4, 5), is y = -40x - 155.