Parabola exercise and checking a log-equation

Hi !

Quote:

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).

$ax^2+ax+1=0$

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

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

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

$\Delta=a^2-4a$

$\implies a(a-4)=0 \implies a=... \ or \ a=...$
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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 :
$y=f'(m)(x-m)+f(m)$

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

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

Quote:

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

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

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

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

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

$ 2x+7 = 125x $

$ x = 118 $

(Nod)

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