# Tutor profile: Coenrad V.

## Questions

### Subject: Pre-Calculus

Explain why the graph of the function f(b) = tan b rises sharply as it approaches 90º, and why it falls sharply as it approaches -90º in the opposite direction.

Consider a unit circle with its centre at the origin of the Cartesian plane. A right-angled triangle can be drawn on any point along the circle so that the line from the origin to the circumference is the hypotenuse. Drawing a perpendicular from the point along the circle to the x-axis creates a triangle with sides r = 1 (hypotenuse), y (vertical perpendicular from x-axis to edge of circle) and x (horizontal line along x-axis from origin to vertical perpendicular y). Now, consider the angle b between the hypotenuse r, and the horizontal side x. We know that tan b = y/x, so for b = 0º, we have y = 0 and x = 1 so that tan0º = 0/1 = 0. As we move counter-clockwise along the circle on the Cartesian plane, the angle will increase from 0º to 90º. Drawing a triangle at each point as we move along the circle, we see that the length of vertical side y increases from 0 to 1, and the length of horizontal side x decreases from 1 to zero. But, as we approach 90º, x becomes infinitely smaller, while y approaches the value 1. That means that tan b = y/x approaches infinity as y approaches 1 and x approaches zero. Therefore, there is a steep rise in the graph of f(b) = tan b as we approach the asymptote of 90º. The graph will never reach 90º as y/x is undefined where x = 0. In a similar way it can be shown that if we repeat this process, but decrease the angle b from 0º to -90º (moving along the circle in the Cartesian plan in a clockwise direction), the value of tan b will become infinitely small the closer we get to -90º, since the value of y is negative in the fourth quadrant. So as y approaches -1 and x approaches 0, y/x approaches negative infinity.

### Subject: Calculus

Explain how you would determine the nature of the graph that represents the cubic function f(x) = ax³ + bx² + cx + d.

To find the x-intercepts, set ax³ + bx² + cx + d = 0 and solve for x. This can be done using the factor theorem and inspection. The y-intercept is found by setting x = 0, giving the y-intercept as d. Next, the stationary points of the graph can be found by taking the first derivative, which represents the gradient of the curve at any given point. The stationary points will have gradients = 0, so by solving f´(x) = 0 for x, we find the two x components of the coordinates of the stationary points. Substituting these into the original function, we find the corresponding y-values of the stationary point coordinates. Finding the second derivative and determining where f´´(x) is positive and negative, will tell us where the graph is concave up (f´´(x) > 0) and concave down (f´´(x) < 0). From this we can then deduce whether a stationary point is a turning point, as well as whether it is a local maximum or local minimum. Where f´´(x) = 0, we have a point of inflection (the point at which the graph changes in concavity – from concave up to concave down, or the other way around). Combining the coordinates of the x-intercepts, the y-intercept, the coordinates of the stationary points and where the graph is concave up or down, we can then sketch the graph, based on this information.

### Subject: Algebra

Explain the different methods by which the roots of a function f(x) = ax² + bx + c can be found, and comment on the suitability of each method.

The easiest way to solve the equation is to factorize the equation into the form (rx + s)(ux + v) = 0 and to solve each component rx + s = 0 and ux + v = 0. This method is useful if you can find obvious combinations where r, s, u and v are integers. Another method is to use the quadratic formula x = (-b ± √ (b² - 4ac)) / 2a. This method is particularly useful when the equation cannot be factorized in the form (rx + s)(ux + v) = 0, and where the roots are irrational or imaginary (non-real). Another advantage of the quadratic formula is that it can be used to determine whether roots are real, non-real, rational or irrational, without having to solve the roots. This can be done by looking at the discriminant: b² - 4ac. A third method to find the roots is to complete the square. This can be done by rewriting the equation in the form a(x² + p) + q = 0 and solving the roots as x = -p ± √ (-q/a). Although similar to using the quadratic method (the quadratic formula is derived by completing the square of the general quadratic equation), writing the equation in the form a(x² + p) + q has the advantage of giving us the turning point of the parabola as (-p;q)

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