Simplify the expression. 18+12x.
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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: