# How can I find the quadratic equation from highest point on parabola? [closed]

If I have a parabola, where the vertex is in $P=(2,0)$, and a point on the parabola is $Q=(-1,-6)$, how can I find the quadratic equation of the form:

$$f(x) = ax^2 + bx + c$$

Vertex form of a quadratic is: $f(x)=a(x-h)^2+k$, where $(h,k)$ is the vertex. So, in our case the vertex is $(2,0)$ so $h=2$ and $k=0$. So we have, $f(x)=a(x-2)^2$. Now, find $a$ by using the point $(-1,-6$). So we have, $-6=a(-1-2)^2$ which gives us $-6=9a$ so $a=-\frac{2}{3}$. So our equation is $f(x)=-\frac{2}{3}(x-2)^2$. Now, multiply this out to get $$f(x)=-\frac{2}{3}x^2+\frac{8}{3}x-\frac{8}{3}$$

Hint: solve $\,f(2)=0\,, \;f'(2)=0\,, \;f(-1)=-6\,$ for $\,a,b,c\,$.

• Something tells me calculus is out of scope for this question... Commented Dec 16, 2016 at 18:26
• @ChocolateAndCheese Maybe, maybe not. That's precisely why a well-asked question should provide some context and OP's own take on it. The question here did neither.
– dxiv
Commented Dec 16, 2016 at 18:33

Here is another approach. Since the parabola is symmetric about the axis of symmetry, we know that $(5, -6)$ is another point on the parabola. Since $f(5)=-6=f(-1)$ and $f(2)=0$, it follows that \begin{align} 4a + 2b + c &= 0\\ a-b+c &= -6\\ 25a+5b+c&=0 \end{align} which you can solve (for example by Gaussian elimination) to get $a=-2/3, b=8/3$ and $c=-8/3$.

• Start by using the "shifted" or "vertex" form $f(x) = a(x - h)^2 + k$; since you know the vertex, you know everything but $a$.

• You can find $a$ since you know $f(-1) = -6$; set up and solve that equation for $a$.

Then just multiply everything out to get the form you're after. (Also the title is a bit misleading; there's only a "highest point" on a parabola if it opens downward, and in this case, the highest point is the vertex.)

$$(x_v,y_v)=\left(-\frac{b}{2a},-\frac{\Delta}{4a}\right)=(2,0)$$

$b=-4a$ and $\Delta=0 \Rightarrow b^2=4ac$.

$$16a^2=4ac \Rightarrow 4a(4a-c)=0$$

But, $a\ne 0$ so $c=4a$.

$$f(x)=a(x^2-4x+4)=a(x-2)^2$$

But $f(-1)=-6$ so $a=-2/3$

$$f(x)=-\frac{2}{3}(x-2)^2$$

The vertex is the extreme point, at which $f'(x) = 0$.

Since $f'(x) = 2ax+b$, this is zero when $x = -\frac{b}{2a}$. At this point

$\begin{array}\\ f(x) &=a(-\frac{b}{2a})^2+b(-\frac{b}{2a})+c\\ &=\frac{ab^2}{4a^2}-\frac{b^2}{2a}+c\\ &=\frac{b^2}{4a}-\frac{b^2}{2a}+c\\ &=-\frac{b^2}{4a}+c\\ \end{array}$

Since this point is $(2, 0)$, $-\frac{b}{2a}=2$ and $-\frac{b^2}{4a}+c = 0$, or $b = -4a$ and $c = \frac{b^2}{4a} =\frac{16a^2}{4a} =4a$.

Since the parabola also passes through $(-1, -6)$, we get $-6 = a(-1)^2+b(-1)+c = a-b+c =a-(-4a)+4a =9a$ or $a=-\frac{6}{9}=-\frac{2}{3}$.

$b$ and $c$ follow.

• Thanks for the correction. Commented Dec 16, 2016 at 23:38