Derivation of $y=a(x-h)^2+k$ from $y=ax^2+bx+c$ given a vertex and a point

Derive $y=a(x-h)^2+k$ from $y=ax^2+bx+c$ given a vertex and a point.

Recently I have been solving a problem to which I could not find a solution. Google search of "quadratic equation given vertex and a point," yielded what I have been looking for. However, although I have already solved the problem, I am still wondering how to derive the equation I was looking for before I googled it.

I was given a vertex $V(-3, -2)$ and a point $P(-4, 0)$ of a parabola. Using $y=ax^2+bx+c$, I have derived an equation for finding the vertex for each parabola with $V(\frac{-b}{2a}, -a(\frac{b}{2a})^2+c)$.

I knew that given the same vertex, the parabola $y=x^2+6x+7$ was close to what I've been looking for. However, this parabola didn't go through $P(-4, 0)$, because it was too wide.

At this point it seemed as if I had enough information to derive an equation, that given a vertex and a point I would obtain a parabola according to the restrictions. However, from this point on I didn't know how to proceed further.

• Note that by symmetry, $(-2,0)$ is also a point on the parabola. – user137731 Jul 31 '16 at 23:53
• Is your goal to determine an equation of the parabola that has vertex $V(-3, -2)$ and passes through the point $P(-4, 0)$? – N. F. Taussig Jul 31 '16 at 23:56
• @N.F.Taussig Yes. – user270346 Jul 31 '16 at 23:56
• I was looking for a general equation, that given the vertex and a point of a parabola I could determine it's coefficients $a, b, c$ as in $y=ax^2+bx+c$. – user270346 Jul 31 '16 at 23:57
• Use the two points you're given and the one you get from symmetry. Plug each in to get a system of linear equations that you can hopefully then solve. – user137731 Jul 31 '16 at 23:58

$$y = ax^2 + bx + c$$

$$y - c = a\left(x^2 + \frac{b}{a}x\right)$$

$$y - c + a\frac{b^2}{4a^2}= a\left(x^2 + \frac{b}{a}x + \frac{b^2}{4a^2}\right)$$

$$y - c + \frac{b^2}{4a}= a\left(x+\frac{b}{2a}\right)^2$$

$$y = a\left(x-\left(-\frac{b}{2a}\right)\right)^2 + \left(c- \frac{b^2}{4a}\right)$$

$$y = a\left(x-h\right)^2 + k$$

where $h = -\frac{b}{2a}$ and $k = c- \frac{b^2}{4a}$

If you're going in reverse (going from vertex $(h,k)$ and point $(x,y)$ to quadratic parameters $a,b,c$), then you can take the last few equations, isolate $a,b,c$, and translate them into terms of $h,k,x,y$:

$$a = \frac{y-k}{(h-x)^2}$$

$$b = -2ha = \frac{-2h(y-k)}{(h-x)^2}$$

$$c = k + \frac{b^2}{4a} = k + \frac{h^2(y-k)}{(h-x)^2}$$

• This answer followed by the @N. F. Taussig answer is the order what makes most sense to me after picking up on this problem from where I left it. Thanks! – user270346 Aug 1 '16 at 0:35
• @Marcus Stuhr, Regarding your edit, note that the parabola $y=x^2+6x+7$ doesn't have roots $x=-4$ and $x=-2$ as required. – user270346 Aug 1 '16 at 0:47
• I've done as you've requested. Please, take your time. Thanks. – user270346 Aug 1 '16 at 0:55
• @Matt I believe you're looking for $2x^2 + 12x + 16 = 0$, which goes through $(-4,0)$ and $(-2,0)$ and has vertex $(-3,-2)$. – Marcus Andrews Aug 1 '16 at 1:15

Since you know that the parabola has vertex $V(-3, -2)$, the vertex form of its equation is $$y = a[x - (-3)]^2 + (-2) = a(x + 3)^2 - 2$$ Since the parabola passes through the point $P(-4, 0)$, we can substitute $-4$ for $x$ and $0$ for $y$ to determine $a$. \begin{align*} 0 & = a(-4 + 3)^2 - 2\\ 2 & = a(-1)^2\\ 2 & = a \end{align*} Hence, $y = 2(x + 3)^2 - 2$. You can expand this expression to obtain the standard form of the equation.

Addendum: In general, if you know the vertex $V(h, k)$ and a point $P(u, v) \neq V(h, k)$ on the parabola, you can write $$y = a(x - h)^2 + k$$ then substitute $u$ for $x$ and $v$ for $y$ to determine $a$. \begin{align*} v & = a(u - h)^2 + k\\ v - k & = a(u - h)^2\\ \frac{v - k}{(u - h)^2} & = a \end{align*} Then $$y = a(x - h)^2 + k = \frac{v - k}{(u - h)^2}(x - h)^2 + k$$ Again, expanding the expression to obtain its standard form enables you to determine the coefficients $a$, $b$, and $c$.

• Thanks for your answer, however that's not my question. Please forgive me if I was not clear enough. I was looking for a general equation, that given the vertex and a point of a parabola I could determine it's coefficients $a, b, c$ as in $y=ax^2+bx+c$, based on what I knew. So, it's more about how do you derive the general equation for solving this type of problems, rather than solving the problem itself. – user270346 Aug 1 '16 at 0:04
• This is especially useful due to the other point suggested by the symmetry. Thanks for your answer. – user270346 Aug 1 '16 at 0:37
• On a side note, do you have any idea as why the vertex equation for this parabola $y=2(x+3)^2-2$ and $y=x^2+6x+7$ following from it, don't have same roots? – user270346 Aug 1 '16 at 0:51
• If you expand $y = 2(x + 3)^2 - 2$, you obtain \begin{align} y & = 2(x + 3)^2 - 2\\ & = 2(x^2 + 6x + 9) - 2\\ & = 2x^2 + 12x + 18 - 2\\ & = 2x^2 + 12x + 16 \end{align} You forgot to multiply the terms in the parentheses by $2$. As you can verify, both $y = 2(x + 3)^2 - 2$ and $y = 2x^2 + 12x + 16$ have the roots $-4$ and $-2$. – N. F. Taussig Aug 1 '16 at 1:00

Generalized Form:

Given $y=ax^2+bx+c$, we can move the loose number $c$ to the other side and try completing the square! So from that, we get $$y-c=ax^2+bx\tag{1}$$

Factoring out $a$ gives us $y-c=a\left(x^2+\frac bax\right)$. Completing the square by dividing the coefficient of the $x$ term by $2$ and squaring it gives us $$y-c+a\left(\frac {b^2}{4a^2}\right)=a\left(x+\frac {b}{2a}\right)^2\tag{2}$$

Simplifying the left hand side and move it to the right hand side to obtain $$y=a\left(x-\left(-\frac {b}{2a}\right)^2\right)+\left(\frac {4ac-b^2}{4a}\right)\tag{3}$$

And since the Vertex follows the formula $(h,k)=\left(-\frac {b}{2a},\frac {4ac-b^2}{4a}\right)$, you get the Vertex formula by plugging it in.