Dear friends,
The function g is defined by g(x)= , x not equal to 3 and 5.
i) Find the coordinates of the turning point of the curve y=g(x).
ii) Hence, sketch the graph of y=g(x) and write down the range of the function g.
Dear friends,
The function g is defined by g(x)= , x not equal to 3 and 5.
i) Find the coordinates of the turning point of the curve y=g(x).
ii) Hence, sketch the graph of y=g(x) and write down the range of the function g.
Turning points occur where the derivative is .
So you need to differentiate the function, set it equal to and solve for . Then back-substitute to find the corresponding .
To differentiate this function, you need to rewrite it as
.
You will now need to use a combination of the Chain Rule and the Product Rule (or expand the inside and just use the Chain Rule).
*Ahem* Attention all posters (including the original poster):
Since this question is posted in the PRE-calculus subforum, the expectation is that the question is to be done without using calculus.
Draw the graph of the quadratic function y = (x - 5)(x + 3). Now think about what happens to this graph when you take the reciprocal of the y-coordinates for each x-coordinate.
The black parabola is the graph of
The pink asymptotic curve is the graph of
The y-axis is not suitably scaled to effectively view the pink curve.
If we "stretch" it, you'd more easily see that the local maximum (turning point) of the pink graph
occurs at the exact same x which locates the minimum value (turning point) of the U-shaped parabola.
If you invert a quadratic equation, the zeros go towards infinity (become vertical asymptotes),
local maxima become local minima and local minima become local maxima.