# 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 $c$?

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Hint: how is the vertex defined and how the two intersection points? – al-Hwarizmi Jul 28 '13 at 13:04

Knowing the vertex, $(x_0, y_0)$ of a parabola tells us the equation of the parabola can be written as:

$$(y - y_0) = p(x - x_0)^2$$

$$y - (-169) = p(x - 13)^2$$ $$y + 169 = p(x^2 - 26 x + 169)$$ $$y = p(x^2 - 26x + 169) - 169$$

Now use the fact that a "vertical parabola" (of the form $f(x) = y = ax^2 + bx + c$) intersects the $y$ axis at points which are symmetrical about the line $x = x_0$. For this parabola, we know the vertex is $(x_0, y_0) = (13, -169)$, so $x = x_0 = 13$ is the line of symmetry. Given that the distance between the $x$-intercepts is $d = 26$, the "roots" of the equation $y = 0 = p(x^2 - 26x + 169) - 169$ must be $x_1 = 13 - 13 = 0, x_2 = 13 + 13 = 26$, and solving for $p$ then gives us $p = 1$.

This gives us that $$f(x) = y = p(x^2 - 26x + 169) - 169 \iff f(x) = x^2 - 26x \implies a = 1, b = -26, c = 0$$

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Clear and concise +1 – Amzoti Jul 29 '13 at 0:59

Use the vertex form of the quadratic:

$$f(x)=m(x-h)^2+k$$

Where the vertex is $(h,k)$.

\begin{align}f(x)&=m(x-13)^2-169 \\ f(x)&=m(x^2-26x+169)-169 \\ f(x)&=mx^2-26mx+(169m-169) \end{align}

The distance between zeros on the x-axis of a polynomial in standard form $x^2+bx+c$ is:

$$d=\frac{\sqrt{b^2-4ac}}{a}$$

Substitute in $a=m$, $b=26m$, $c=169m-169$, and $d=26$, and solve for $m$.

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There are various ways of solving this. Perhaps the easiest is to note that the "vertex" is the place where the vertical line $x=13$ is a line of symmetry.

The zeros of the quadratic are placed symmetrically either side of this line - so $13$ away on each side, to make the distance between them equal to $26$ - so zeros are $x=0, 26$.

Any quadratic function with these zeros has form $a(x-0)(x-26)=ax(x-26)$

And it should be easy to solve from there.

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$$y=ax^2+bx+c \implies \left(x+\frac b{2a}\right)^2=4\cdot\frac14\left(y-\left(c-\frac{b^2}{4a}\right)\right)$$

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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