# Thread: Parabola exercise and checking a log-equation

1. ## Parabola exercise and checking a log-equation

The parabola exercise is this one:

'A parabola is given by the equation y = ax^2 + ax + 1, where a is a positive real number different from zero.
(a): Determine a so that the parabola has exactly one intersection point with the x-axis.
(b): At the point P(x0,y0) the parabola has a tangent which is parallel to the line with the equation y = 8x - 800. Find the value of X0'

And then I had this log equation, and if someone would check if it's right, that'd make me very happy. (:

$\displaystyle \log_5 (2x+7) - 3 = \log_5 x$

$\displaystyle \log_5 (2x+7) - \log_5 125 = \log_5 x$

$\displaystyle \log_5 ((2x+7)/125) = \log_5 x$

$\displaystyle (2x+7)/125 = x$

$\displaystyle 2x+7 = 125x$

$\displaystyle x = 118$

2. Hi !

Originally Posted by Fnus
The parabola exercise is this one:

'A parabola is given by the equation y = ax^2 + ax + 1, where a is a positive real number different from zero.
(a): Determine a so that the parabola has exactly one intersection point with the x-axis.
(b): At the point P(x0,y0) the parabola has a tangent which is parallel to the line with the equation y = 8x - 800. Find the value of X0'
Question a).

$\displaystyle ax^2+ax+1=0$

A quadratic equation has one and only one solution if it is proportional to a perfect square.

This means that you want $\displaystyle ax^2+ax+1$ to be a perfect square.

The discriminant $\displaystyle \Delta$ should be equal to 0.

$\displaystyle \Delta=a^2-4a$

$\displaystyle \implies a(a-4)=0 \implies a=... \ or \ a=...$
----------------------------

Question b)
two parallel lines have the same slope. Here, the slope is 8.

A tangent line to the curve representing a function f(x) has this equation :
$\displaystyle y=f'(m)(x-m)+f(m)$

So actually, you have to solve for $\displaystyle f'(m)=8$.

Is it clear enough ? Because I didn't understand what X0' was...

And then I had this log equation, and if someone would check if it's right, that'd make me very happy. (:

$\displaystyle \log_5 (2x+7) - 3 = \log_5 x$

$\displaystyle \log_5 (2x+7) - \log_5 125 = \log_5 x$

$\displaystyle \log_5 ((2x+7)/125) = \log_5 x$

$\displaystyle (2x+7)/125 = x$

$\displaystyle 2x+7 = 125x$

$\displaystyle x = 118$

Just check, in this kind of exercises, that the answer fits in the domain of the equation.

that is to say 2x+7>0 and x>0