I know how to find the parabola quadratic equation given the roots. However, in this problem I'm given the y-intercept of (0,3) {So, now I know C value} and the axis of symmetry of x=-3/8. From this I know that 3a=4b. But, it seems I need one more point, since I have 3 unknowns, I'd need 3 INDEPENDENT equations.

So, I said, ah, I know another point, it's on the other side of the aos. So, I used (-3/4, 3). However, when I plug this in, it appears to not be an independent equation. So, I'm left with just c=3 and 3a=4b. How, do I work this from here?


The point you propose to use, (-3/4,3) just follows from symmetry of the parabola about x = -3/8. So, you are not actually fixing a degree of freedom of the parabola. In a sense, the parabola can slide up and down. Let me give 2 examples, to illustrate. Consider, y = $64(x-3/8)^2/3$ which follows all the properties above, and y = $64(x+3/8)^2-6$ which again satisfies the properties. So, the solution to the question will not be a parabola, it will be a family of parabolas, $y_t(x) = t(x+3/8)^2+3-9t/64$.


The symetry axis gives the point where the derivative is zero.

let $y=f(x)=ax^2+bx+c$ be the equation of the parabola.

the derivative is $2ax+b$ and gives

$2a\frac{-3}{8}+b=0$ or


the point $(0,3)$ satisfy the condition


finally, we get the expression


you have an infinite number of solutions. try it graphically.

  • $\begingroup$ So, the problem as given is not solvable? The book gave an answer of 4x^2 +3x +3=0, which of course does work. I see that 8x^2 +6x +3=0 also works. I guess the book was giving a "possible" answer by just letting a=4. They could have let a = any value. I see this now. Thanks. $\endgroup$ – user163862 Oct 13 '16 at 15:54

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