An equation of the line tangent to the graph of f(x)=(4x^2-8x+3)^4 at the point where x=-1 is
To find the equation of the line tangent to the graph of f(x) at the point where x=-1, we first need to find the slope of the tangent line.
The slope of a tangent line to a curve can be found by taking the derivative of the function at that point.
First, let's find the derivative of f(x):
f'(x) = 4(4x^2-8x+3)^3 * (8x-8)
Now, let's find the slope of the tangent line when x=-1:
f'(-1) = 4(4(-1)^2-8(-1)+3)^3 * (8(-1)-8)
= 4(4+8+3)^3 * (-8-8)
= 4(15)^3 * (-16)
= -153,600
So, the slope of the tangent line when x=-1 is -153,600.
Next, we can use the point-slope form of a line to find the equation of the tangent line.
Using the point (-1, f(-1)), we have:
y - f(-1) = m(x - (-1))
y - f(-1) = -153,600(x + 1)
Finally, let's simplify the equation:
y - f(-1) = -153,600x - 153,600
The equation of the line tangent to the graph of f(x)=(4x^2-8x+3)^4 at the point where x=-1 is:
y = -153,600x - 153,600 + f(-1)