
To find the gradient of the tangent, we must first differentiate the function.

The gradient at x = 0 is given by evaluating f'(0).

The derivative of the function at this point is negative, which tells us <em>the function is decreasing at that point</em>.
The tangent to the line is a straight line, so we will have a linear equation of the form y = mx + c. We know the gradient, m, is equal to -1, so

Now we need to substitute a point on the tangent into this equation to find c. We know a point when x = 0 lies on here. To find the y-coordinate of this point we need to evaluate f(0).

So the point (0, -1) lies on the tangent. Substituting into the tangent equation:
Answer:
Check the attachment, then plug in values as needed
Step-by-step explanation:
Answer:
none of the given options is true
Step-by-step explanation:
Given: u=(-8,8) , v= (-1,2)
To find: vectors
such that 
Solution:
A vector is a quantity that has both magnitude and direction.

Let 
So,

So, u = 
So, none of the given options is true
Answer:

Step-by-step explanation:

Ok so the first thing you would do is input the number where The variables are in the equation so 8^2-5(4)/4