# Thread: Parabola and tagent of the x-axis

1. ## Parabola and tagent of the x-axis

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.

$\displaystyle y=4x^2+bx+25$

2. 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.

3. thanks, i didnt see the problem like that. sorry for the wording, i had to translate it from french. Im not to familiar with the english terms for math for each one of them.

4. Originally Posted by DAngel
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.
$\displaystyle y=4x^2+bx+25$
Simply put: for tangency to exist, sqrt(b^2 - 4ac) = 0;
so b^2 = 4ac
b^2 = 4(4)(25) : b = 20

C'est bien? J'espere que oui.

5. 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