i have this question where i need to find the value of b where the formula of a parabola is tagent to the x-axis but i have no idea how to proceed.
$\displaystyle y=4x^2+bx+25$
thanks in advance.
The wording of your question was confusing. I think you are trying to find the value of b that makes the stated parabola tangential to the x axis.
You need the curve to intersect the x axis and have a gradient of 0 at the point of intersection. Since you posted in the pre calculus part of the forum I assume you aren't able to use calculus to get a pair of simultaneous equations to solve.
Instead you'll have to use the properties of the graph. A parabola has only 1 turning point. If it is tangential to the x-axis then it cant cross it. taken together this means that the equation f(x) = 0 has only 1 solution, ie the graph has a double root.
If a parabola has a double root it must be possible to factorise it as follows:
$\displaystyle y=(ax + c)^2$
expanding the brackets:
$\displaystyle y=a^2x^2 + 2ac +c^2$
you know that the equation is
$\displaystyle y=4x^2 + 2b +25$
comparing coefficients:
coefficients of $\displaystyle x^2$
$\displaystyle a^2 = 4 \rightarrow a=2$
coefficients of $\displaystyle x^0$(constant term)
$\displaystyle c^2=25 \rightarrow c = 5$
coefficients of $\displaystyle x$
b=2ac
b= 20
in case you dont believe me, the graph is plotted here:
http://www.wolframalpha.com/input/?i=plot+4x^2+%2B20x+%2B25
You can see it is tangential to the x axis as required.
NB: the above solution is not unique. b=-20 works too. You can get this by taking the negative square root instead of the positive one when comapring coefficients.
The minimum value of a quadratic function f(x)=ax^2 + bx + c occurs at x=-b/2a
therefore:
x= b/8
setting y=0
0=4(-b/8)^2 + b(-b/8) + 25
therefore
0= b^2/16 - 2b^2/16 +25
b^2/16 = 25
b=20
Sorry... new here and have still to decipher how to use the equation editor