Equation of parabola given 2 points $(x_1,y_1)$ and $(x_2,y_2)$ in expanded form

Question

I need to find an equation for the parabola that passes through the points $(0,0)$ and $(5,0)$, such that $f(x)<0$ whenever $0< x <5$. The answer should be in expanded form. I.e., $f(x)=ax^2+bx+c$.

I have tried to substitute the two points into the equation of the parabola $y = \frac{1}{2(b+k)}{(x-a)^2}+\frac{(b+k)}{2}$. But the problem I am facing is that there are 3 variables(a,b,k) here and only 2 equations. I have also tried substituting the point in $f(x)=ax^2+bx+c$. I am just starting with parabolas so any book or resource on this topic would be greatly appreciated. Thank you in advance.

score 1 · Accepted Answer · answered May 17 '16 at 05:14

1

If you were clever you would say, I know 2 roots, therefore: $y = a(x-0)(x-5)$ now, what values can $a$ take on?

If you were more brute force about it.

$f(x) = ax^2 + bx + c\\ f(0) = c = 0\\ f(5) = 25a + 5b = 0\\ b = -5a\\ f(x) = ax^2 -5a x\\ $

And you still need to make sure that a is in the right set of values.

but you have chosen

$y = a (x-h)^2 + v$

This is probably not the right place to start. But there is enough information given to determine that $h = \frac 52$, (halfway between the roots) and the sign of $a.$ Still, not the approach I would recommend.

answered May 17 '16 at 05:14

Doug M

57,877

Now I understand it , after finding $f(x) = ax^2-5ax$, you find the variable 'a' using the quadratic formula. Thank you – Blacq_Shadow May 18 '16 at 05:14
So the answer would be $f(x) = x^2 - 5x +0$. – Blacq_Shadow May 18 '16 at 05:20

score 0 · Answer 2 · answered May 17 '16 at 05:18

You do not have enough information to get an explicit value for a.

As you said, your function will have the form:

$$y=ax^2+bx+c$$ Plugging in $y=0$ and $x=0$ tells us that $0=c$. So now we have the new equation: $$y=ax^2+bx$$ Plugging in the second point we see that $0=25a+5b\Rightarrow 5a=-b$ yielding now

$$y=ax^2-5ax$$ And now we need to use your last condition, that $f(x)<0$ whenever $0<x<5$.

So when is $y=f(x)=ax^2-5ax<0\Rightarrow a(x^2-5x)<0$.But between 0 and 5, $x^2-5x<0$, which you can verify by plugging in some values, so $a>0$ in order to not change the sign of $f(x)$.

score 0 · Answer 3 · answered May 17 '16 at 07:03

Three points determine a parabola with axes parallel to either x- or y-. One more point or condition is lacking here for full a solution. But upto a constant,

$ y = a \cdot x \cdot (5-x) $ are all parabolas with variable $a$ passing through the given points.

Equation of parabola given 2 points $(x_1,y_1)$ and $(x_2,y_2)$ in expanded form

3 Answers3