Question

Let f be the function given by f(x) = y = -x^2 + 9(pi/2)^2, the graph of f crosses the y-axis at point P and the x-axis at point Q.

a) Write an equation for the line passing through the point P and Q
b) Write an equation for the line tangent to the graph of f at point Q. Show the analysis that leads to your conclusion.
c) Find the x-coordinate of the point on the graph of f, between P and Q at which the line tangent to the graph of f is parallel to line PQ.

Answer 1

Put x=0 and y=0 respectively to find the coordinates of the point P and Q.After this you will be able to compute the equation of the line ...

Answer 2

the only trick in the problem is the introduction of pi which is probably to make the calculation a bit scary .. :P

Answer 3

For y-intercept, x=0
\[y = \frac{9\pi^2}{4}\]
For x-intercept y=0

\[x = \pm \frac{3\pi}{2}\]
Since, the equation is \(y = -x^2 + \frac{9\pi^2}{4}\) we know that it's a parabola opening downwards. You can form an equation now using \[y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}x - x_1\]

Now you can calculate the slope by differentiating the curve at point Q for the tangent
Let's assume at point Q oridnate is \(\frac{3\pi}{2}\)
\[\frac{dy}{dx} = -2x \]
\[\left. \frac{dy}{dx}\right|_{x = \frac{3\pi}{2}} = -3 \pi \]

For equation of the tangent

\[y - y_1 = \frac{dy}{dx}|_{x = \frac{3\pi}{2}} (x - x_1)\]

At Point Q \(x = \frac{3\pi}{2}\) and y is zero

Answer 4

@Ishaan94: Great answer, but this is probably Homework so .. :/

Answer 5

Oh yeah it is homework

@hmm I have done only first two parts of the questions, you should do the third one on your own