The zeros of a quadratic relation are $0$ and $6$. The relation has a minimum value of $-9$. Find an equation of the parabola.

Hello, I've tried to come up with a solution to the exercise but it's complicated. I don't really know if it's right or not.

The quadratic equation is: $ax^2+bx+c=0$

Then we have $3$ points: $(0,0),(0,6),(x,-9)$ \begin{align*} c & = 0\\ 36a+6b+c & = 0\\ -b^2+4ac+36a & = 0 \end{align*} The $y$-coordinate of the vertex is $-\Delta/4a$.

Then we will use $c$ and the substitution method to solve $a,b,c$. Is that right?

Thank you for your help.

  • $\begingroup$ Please read this tutorial on how to typeset mathematics on this site. $\endgroup$ – N. F. Taussig Aug 22 '17 at 19:08

If by a "quadratic relation" you mean $y=ax^2+bx+c,$ then you have $y = a(x-0)(x-6),$ so that $y=0$ if $x$ is either $0$ or $6.$ Then you need to figure out which value of $a$ will make the minimum $y$-value $-9.$ Symmetry shows that if the $x$-intercepts are $0$ and $6$ then the minimum $y$-value occurs when $x$ is halfway between $0$ and $6$. So $y = a(3-0)(3-6) = -9,$ so what is $a$?

  • $\begingroup$ oh your way is much easier to understand. Thank you so much! $\endgroup$ – KhanhNguyen Aug 22 '17 at 18:36
  • $\begingroup$ @KhanhNguyen : I'm glad it helps. $\endgroup$ – Michael Hardy Aug 22 '17 at 18:39

If the roots are $0$ and $6$, the quadratic equation will have the form

$$y=f (x)=a (x-0)(x-6)=a (x^2-6x) $$

the minimum is attained at $c $ such that $$f'(c)=a (2c-6)=0$$ thus $c=3$ and $$f (3)=-9a=-9$$ which gives $a=1$

finally the desired equation is $$y=x (x-6) $$ $$=(x-3)^2-9\ge -9$$


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