Science:Math Exam Resources/Courses/MATH110/December 2011/Question 07 (e)/Solution 1

• $f(x)=x^{2}-6x$

As suggested in the hint, first find the roots of the function.

$x^{2}-6x=0$

$x(x-6)=0$

$f(x)$ has roots at x = 0 and x = 6. Using this fact, or by calculating where the derivative of $f$ is zero, we can deduce that the minimum of $f(x)$ is at x = 3. To find the y-value at this point we simply plug in x = 3 to get $y=(3)^{2}-6(3)=-9$. Plotting the roots and the minimum on the graph should allow you to sketch the curve.

• $P=(4,-12)$

• Tangent lines that pass through P

We know there are two lines. Both pass through the point $(4,-12)$ and their two slopes are -2 and 6 (from part c). The equation of the two lines are thus

$y+12=-2(x-4)$

$y+12=6(x-4)$

Or in y = mx + b form

$y+12=-2x-4$

$y=6x-36$

Alternatively, you know that the line with slope -2 passes through P and the point (2, f(2)) = (2, -8), while the line with slope 6 passes through P and the point (6, f(6)) = (6, 0). Drawing these points and then connecting them with lines will also produce the correct picture.