<s>[INST] How to search quadratic function If a graph of the quadratic function $f(x)=ax^2+bx+c$, where $a$, $b$ and $c$ are constants. If this function vertex is $(13,−169)$ and the distance between the two intersection points with the $x$-axis is $26$, what is the value of $a$, $b$, and $c$?
 [/INST] $$y=ax^2+bx+c \implies \left(x+\frac b{2a}\right)^2=4\cdot\frac14\left(y-\left(c-\frac{b^2}{4a}\right)\right)$$
Comparing with $(x-h)^2=4a(y-k)$  the vertex will be $\left(-\frac b{2a},c-\frac{b^2}{4a}\right)$
So, $-\frac b{2a}=13\implies b=-26a\ \ \ \ (1)$
and $c-\frac{b^2}{4a}=-169\implies b^2-4ac=169\cdot4a\ \ \ \ (2)$
$\implies 4ac=b^2-169\cdot4a=(-26a)^2-169\cdot4a$
$\implies c=169(a-1) \ \ \ \ (3)$ assuming $a\ne0$
For the  two intersection points with the x-axis, $y=f(x)=0$
If $(x_1,0),(x_2,0)$ are the  two intersection points, $x_1,x_2$ are the roots of $ax^2+bx+c=0$
$$\implies x_1+x_2=-\frac ba,x_1x_2=\frac ca$$
So, the distance will be $\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}=\sqrt{(x_1+x_2)^2-4x_1x_2}=\frac{\sqrt{b^2-4ac}}{|a|}=\frac{\sqrt{676a}}{|a|}\text{ using }(2)$ 
So, the distance $=26\sqrt a$ assuming $a>0$ otherwise the distance will be complex
Then we have $26\sqrt a=26$
Now use $(1),(3)$ to find $b,c$ respectively
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