To find the equation of the line tangent to the function at the point (-2,6), we need to take the derivative of the function and evaluate it at x = -2.
Given that f(x) = 4x^2 + 5x:
f'(x) = 8x + 5 (derivative of 4x^2 is 8x, derivative of 5x is 5)
Now, evaluate f'(-2):
f'(-2) = 8(-2) + 5 = -16 + 5 = -11
The slope of the tangent at x = -2 is -11. We also know that the tangent line passes through the point (-2,6). We can use the point-slope form of the equation of a line:
y - y1 = m(x - x1)
Here, m is the slope (-11), and (x1, y1) is the point (-2,6). Plugging in the values:
y - 6 = -11(x - (-2))
y - 6 = -11(x + 2)
y - 6 = -11x - 22
y = -11x - 16
Therefore, the equation of the line tangent to the function f(x) = 4x^2 + 5x at the point (-2,6) is y = -11x - 16.
what is the equation of the line tangent to the function f(x)=4x^2+5x at the point (-2,6)?
*y=8x+5
*y=-11x-16
*y=-11x+16
*y=-8x+5
1 answer